complex numbers 1
DESCRIPTION
Basic concepts of Complex numbers ( part 1 )TRANSCRIPT
![Page 1: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/1.jpg)
Complex Numbers
N. B. Vyas
Department of Mathematics,Atmiya Institute of Tech. and Science,
Rajkot (Guj.)
N. B. Vyas Complex Numbers
![Page 2: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/2.jpg)
Definition of Complex Number
Complex Numbers
A number of the form x+ iy, where x and y are
real numbers and i =√−1 is called a complex
number and is denoted by z. It is also denoted
by an ordered pair(x, y).
Thus z = x + iy or z = (x, y)
N. B. Vyas Complex Numbers
![Page 3: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/3.jpg)
Definition of Complex Number
Complex Numbers
A number of the form x+ iy, where x and y are
real numbers and i =√−1 is called a complex
number and is denoted by z. It is also denoted
by an ordered pair(x, y).
Thus z = x + iy or z = (x, y)
N. B. Vyas Complex Numbers
![Page 4: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/4.jpg)
Definition of Complex Number
The set of complex numbers is denoted by {.
If z = x+ iy is a complex number, thenx is called the real part of z and denoted byRe(z).y is called the imaginary part of z and isdenoted by Im(z).
If x = 0 and y 6= 0 then z = 0 + iy = iy
is called a purely imaginary number.
If x 6= 0 and y = 0 then z = x+ i0 = x
is called a real number.
If x = 0 and y = 0 then z = 0 + i.0 = 0is the zero complex number.
N. B. Vyas Complex Numbers
![Page 5: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/5.jpg)
Definition of Complex Number
The set of complex numbers is denoted by {.
If z = x+ iy is a complex number, then
x is called the real part of z and denoted byRe(z).y is called the imaginary part of z and isdenoted by Im(z).
If x = 0 and y 6= 0 then z = 0 + iy = iy
is called a purely imaginary number.
If x 6= 0 and y = 0 then z = x+ i0 = x
is called a real number.
If x = 0 and y = 0 then z = 0 + i.0 = 0is the zero complex number.
N. B. Vyas Complex Numbers
![Page 6: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/6.jpg)
Definition of Complex Number
The set of complex numbers is denoted by {.
If z = x+ iy is a complex number, thenx is called the real part of z and denoted byRe(z).
y is called the imaginary part of z and isdenoted by Im(z).
If x = 0 and y 6= 0 then z = 0 + iy = iy
is called a purely imaginary number.
If x 6= 0 and y = 0 then z = x+ i0 = x
is called a real number.
If x = 0 and y = 0 then z = 0 + i.0 = 0is the zero complex number.
N. B. Vyas Complex Numbers
![Page 7: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/7.jpg)
Definition of Complex Number
The set of complex numbers is denoted by {.
If z = x+ iy is a complex number, thenx is called the real part of z and denoted byRe(z).y is called the imaginary part of z and isdenoted by Im(z).
If x = 0 and y 6= 0 then z = 0 + iy = iy
is called a purely imaginary number.
If x 6= 0 and y = 0 then z = x+ i0 = x
is called a real number.
If x = 0 and y = 0 then z = 0 + i.0 = 0is the zero complex number.
N. B. Vyas Complex Numbers
![Page 8: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/8.jpg)
Definition of Complex Number
The set of complex numbers is denoted by {.
If z = x+ iy is a complex number, thenx is called the real part of z and denoted byRe(z).y is called the imaginary part of z and isdenoted by Im(z).
If x = 0 and y 6= 0 then z = 0 + iy = iyis called a purely imaginary number.
If x 6= 0 and y = 0 then z = x+ i0 = x
is called a real number.
If x = 0 and y = 0 then z = 0 + i.0 = 0is the zero complex number.
N. B. Vyas Complex Numbers
![Page 9: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/9.jpg)
Definition of Complex Number
The set of complex numbers is denoted by {.
If z = x+ iy is a complex number, thenx is called the real part of z and denoted byRe(z).y is called the imaginary part of z and isdenoted by Im(z).
If x = 0 and y 6= 0 then z = 0 + iy = iyis called a purely imaginary number.
If x 6= 0 and y = 0 then z = x+ i0 = xis called a real number.
If x = 0 and y = 0 then z = 0 + i.0 = 0is the zero complex number.
N. B. Vyas Complex Numbers
![Page 10: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/10.jpg)
Definition of Complex Number
The set of complex numbers is denoted by {.
If z = x+ iy is a complex number, thenx is called the real part of z and denoted byRe(z).y is called the imaginary part of z and isdenoted by Im(z).
If x = 0 and y 6= 0 then z = 0 + iy = iyis called a purely imaginary number.
If x 6= 0 and y = 0 then z = x+ i0 = xis called a real number.
If x = 0 and y = 0 then z = 0 + i.0 = 0is the zero complex number.
N. B. Vyas Complex Numbers
![Page 11: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/11.jpg)
Definition of Conjugate Complex Number
Conjugate Complex Numbers
If two complex numbers differ only in the sign of theimaginary part then they are called conjugatecomplex numbers.
Thus x+ iy and x− iy are conjugate complexnumbers.
The conjugate of a complex number z is denotedby z̄.
The conjugate of real number is the real numberitself.
N. B. Vyas Complex Numbers
![Page 12: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/12.jpg)
Definition of Conjugate Complex Number
Conjugate Complex Numbers
If two complex numbers differ only in the sign of theimaginary part then they are called conjugatecomplex numbers.Thus x+ iy and x− iy are conjugate complexnumbers.
The conjugate of a complex number z is denotedby z̄.
The conjugate of real number is the real numberitself.
N. B. Vyas Complex Numbers
![Page 13: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/13.jpg)
Definition of Conjugate Complex Number
Conjugate Complex Numbers
If two complex numbers differ only in the sign of theimaginary part then they are called conjugatecomplex numbers.Thus x+ iy and x− iy are conjugate complexnumbers.
The conjugate of a complex number z is denotedby z̄.
The conjugate of real number is the real numberitself.
N. B. Vyas Complex Numbers
![Page 14: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/14.jpg)
Definition of Conjugate Complex Number
Conjugate Complex Numbers
If two complex numbers differ only in the sign of theimaginary part then they are called conjugatecomplex numbers.Thus x+ iy and x− iy are conjugate complexnumbers.
The conjugate of a complex number z is denotedby z̄.
The conjugate of real number is the real numberitself.
N. B. Vyas Complex Numbers
![Page 15: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/15.jpg)
Properties of Complex Numbers
1 If x+ iy = 0 then x = 0, y = 0
2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y2
3 If x1 + iy1 = x2 + iy2 then x1 − iy1 = x2 − iy2
4 Sum, difference and quotient(division) of any twocomplex numbers is a complex number.
N. B. Vyas Complex Numbers
![Page 16: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/16.jpg)
Properties of Complex Numbers
1 If x+ iy = 0 then x = 0, y = 0
2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y2
3 If x1 + iy1 = x2 + iy2 then x1 − iy1 = x2 − iy2
4 Sum, difference and quotient(division) of any twocomplex numbers is a complex number.
N. B. Vyas Complex Numbers
![Page 17: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/17.jpg)
Properties of Complex Numbers
1 If x+ iy = 0 then x = 0, y = 0
2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y2
3 If x1 + iy1 = x2 + iy2 then x1 − iy1 = x2 − iy2
4 Sum, difference and quotient(division) of any twocomplex numbers is a complex number.
N. B. Vyas Complex Numbers
![Page 18: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/18.jpg)
Properties of Complex Numbers
1 If x+ iy = 0 then x = 0, y = 0
2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y2
3 If x1 + iy1 = x2 + iy2 then x1 − iy1 = x2 − iy2
4 Sum, difference and quotient(division) of any twocomplex numbers is a complex number.
N. B. Vyas Complex Numbers
![Page 19: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/19.jpg)
Properties of Complex Numbers
If z1 = x1 + iy1 and z2 = x2 + iy2 then their
Sum: z1 + z2 = (x1 + iy1) + (x2 + iy2)= (x1 + x2) + i(y1 + y2)
Difference: z1 − z2 = (x1 + iy1)− (x2 + iy2)= (x1 − x2) + i(y1 − y2)
Product: z1.z2 = (x1 + iy1).(x2 + iy2)= (x1x2 − y1y2) + i(x1y2 + x2y1)
N. B. Vyas Complex Numbers
![Page 20: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/20.jpg)
Properties of Complex Numbers
If z1 = x1 + iy1 and z2 = x2 + iy2 then their
Sum: z1 + z2 = (x1 + iy1) + (x2 + iy2)
= (x1 + x2) + i(y1 + y2)
Difference: z1 − z2 = (x1 + iy1)− (x2 + iy2)= (x1 − x2) + i(y1 − y2)
Product: z1.z2 = (x1 + iy1).(x2 + iy2)= (x1x2 − y1y2) + i(x1y2 + x2y1)
N. B. Vyas Complex Numbers
![Page 21: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/21.jpg)
Properties of Complex Numbers
If z1 = x1 + iy1 and z2 = x2 + iy2 then their
Sum: z1 + z2 = (x1 + iy1) + (x2 + iy2)= (x1 + x2) + i(y1 + y2)
Difference: z1 − z2 = (x1 + iy1)− (x2 + iy2)= (x1 − x2) + i(y1 − y2)
Product: z1.z2 = (x1 + iy1).(x2 + iy2)= (x1x2 − y1y2) + i(x1y2 + x2y1)
N. B. Vyas Complex Numbers
![Page 22: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/22.jpg)
Properties of Complex Numbers
If z1 = x1 + iy1 and z2 = x2 + iy2 then their
Sum: z1 + z2 = (x1 + iy1) + (x2 + iy2)= (x1 + x2) + i(y1 + y2)
Difference: z1 − z2 = (x1 + iy1)− (x2 + iy2)
= (x1 − x2) + i(y1 − y2)
Product: z1.z2 = (x1 + iy1).(x2 + iy2)= (x1x2 − y1y2) + i(x1y2 + x2y1)
N. B. Vyas Complex Numbers
![Page 23: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/23.jpg)
Properties of Complex Numbers
If z1 = x1 + iy1 and z2 = x2 + iy2 then their
Sum: z1 + z2 = (x1 + iy1) + (x2 + iy2)= (x1 + x2) + i(y1 + y2)
Difference: z1 − z2 = (x1 + iy1)− (x2 + iy2)= (x1 − x2) + i(y1 − y2)
Product: z1.z2 = (x1 + iy1).(x2 + iy2)= (x1x2 − y1y2) + i(x1y2 + x2y1)
N. B. Vyas Complex Numbers
![Page 24: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/24.jpg)
Properties of Complex Numbers
If z1 = x1 + iy1 and z2 = x2 + iy2 then their
Sum: z1 + z2 = (x1 + iy1) + (x2 + iy2)= (x1 + x2) + i(y1 + y2)
Difference: z1 − z2 = (x1 + iy1)− (x2 + iy2)= (x1 − x2) + i(y1 − y2)
Product: z1.z2 = (x1 + iy1).(x2 + iy2)
= (x1x2 − y1y2) + i(x1y2 + x2y1)
N. B. Vyas Complex Numbers
![Page 25: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/25.jpg)
Properties of Complex Numbers
If z1 = x1 + iy1 and z2 = x2 + iy2 then their
Sum: z1 + z2 = (x1 + iy1) + (x2 + iy2)= (x1 + x2) + i(y1 + y2)
Difference: z1 − z2 = (x1 + iy1)− (x2 + iy2)= (x1 − x2) + i(y1 − y2)
Product: z1.z2 = (x1 + iy1).(x2 + iy2)= (x1x2 − y1y2) + i(x1y2 + x2y1)
N. B. Vyas Complex Numbers
![Page 26: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/26.jpg)
Polar form of a Complex Number
Let P (x, y) be the point which representsz = (x, y) = x+ iy
Let OP = r and ∠POM = θ.Then from ∆OPM .x = OM = rcosθ. y = PM = rsinθ.∴ z = x+ iy = r(cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 27: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/27.jpg)
Polar form of a Complex Number
Let P (x, y) be the point which representsz = (x, y) = x+ iyLet OP = r and ∠POM = θ.
Then from ∆OPM .x = OM = rcosθ. y = PM = rsinθ.∴ z = x+ iy = r(cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 28: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/28.jpg)
Polar form of a Complex Number
Let P (x, y) be the point which representsz = (x, y) = x+ iyLet OP = r and ∠POM = θ.Then from ∆OPM .
x = OM = rcosθ. y = PM = rsinθ.∴ z = x+ iy = r(cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 29: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/29.jpg)
Polar form of a Complex Number
Let P (x, y) be the point which representsz = (x, y) = x+ iyLet OP = r and ∠POM = θ.Then from ∆OPM .x = OM = rcosθ. y = PM = rsinθ.
∴ z = x+ iy = r(cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 30: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/30.jpg)
Polar form of a Complex Number
Let P (x, y) be the point which representsz = (x, y) = x+ iyLet OP = r and ∠POM = θ.Then from ∆OPM .x = OM = rcosθ. y = PM = rsinθ.∴ z = x+ iy = r(cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 31: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/31.jpg)
Polar form of a Complex Number
r is called the absolute value or the modulus ofz and is denoted by |z|.
∴ r = |z| =√x2 + y2 =
√z.z̄
Geometrically, |z| is the distance of point z fromthe origin.
θ is called the argument of z or amplitude of z.
It is denoted by argz or Ampz
∴ θ = argz = tan−1(yx
)θ is the directed angle from the positive X-axis toOP.
The value of θ lies in the interval −π < θ ≤ π iscalled principal value.
N. B. Vyas Complex Numbers
![Page 32: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/32.jpg)
Polar form of a Complex Number
r is called the absolute value or the modulus ofz and is denoted by |z|.∴ r = |z| =
√x2 + y2 =
√z.z̄
Geometrically, |z| is the distance of point z fromthe origin.
θ is called the argument of z or amplitude of z.
It is denoted by argz or Ampz
∴ θ = argz = tan−1(yx
)θ is the directed angle from the positive X-axis toOP.
The value of θ lies in the interval −π < θ ≤ π iscalled principal value.
N. B. Vyas Complex Numbers
![Page 33: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/33.jpg)
Polar form of a Complex Number
r is called the absolute value or the modulus ofz and is denoted by |z|.∴ r = |z| =
√x2 + y2 =
√z.z̄
Geometrically, |z| is the distance of point z fromthe origin.
θ is called the argument of z or amplitude of z.
It is denoted by argz or Ampz
∴ θ = argz = tan−1(yx
)θ is the directed angle from the positive X-axis toOP.
The value of θ lies in the interval −π < θ ≤ π iscalled principal value.
N. B. Vyas Complex Numbers
![Page 34: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/34.jpg)
Polar form of a Complex Number
r is called the absolute value or the modulus ofz and is denoted by |z|.∴ r = |z| =
√x2 + y2 =
√z.z̄
Geometrically, |z| is the distance of point z fromthe origin.
θ is called the argument of z or amplitude of z.
It is denoted by argz or Ampz
∴ θ = argz = tan−1(yx
)θ is the directed angle from the positive X-axis toOP.
The value of θ lies in the interval −π < θ ≤ π iscalled principal value.
N. B. Vyas Complex Numbers
![Page 35: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/35.jpg)
Polar form of a Complex Number
r is called the absolute value or the modulus ofz and is denoted by |z|.∴ r = |z| =
√x2 + y2 =
√z.z̄
Geometrically, |z| is the distance of point z fromthe origin.
θ is called the argument of z or amplitude of z.
It is denoted by argz or Ampz
∴ θ = argz = tan−1(yx
)
θ is the directed angle from the positive X-axis toOP.
The value of θ lies in the interval −π < θ ≤ π iscalled principal value.
N. B. Vyas Complex Numbers
![Page 36: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/36.jpg)
Polar form of a Complex Number
r is called the absolute value or the modulus ofz and is denoted by |z|.∴ r = |z| =
√x2 + y2 =
√z.z̄
Geometrically, |z| is the distance of point z fromthe origin.
θ is called the argument of z or amplitude of z.
It is denoted by argz or Ampz
∴ θ = argz = tan−1(yx
)θ is the directed angle from the positive X-axis toOP.
The value of θ lies in the interval −π < θ ≤ π iscalled principal value.
N. B. Vyas Complex Numbers
![Page 37: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/37.jpg)
Polar form of a Complex Number
r is called the absolute value or the modulus ofz and is denoted by |z|.∴ r = |z| =
√x2 + y2 =
√z.z̄
Geometrically, |z| is the distance of point z fromthe origin.
θ is called the argument of z or amplitude of z.
It is denoted by argz or Ampz
∴ θ = argz = tan−1(yx
)θ is the directed angle from the positive X-axis toOP.
The value of θ lies in the interval −π < θ ≤ π iscalled principal value.
N. B. Vyas Complex Numbers
![Page 38: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/38.jpg)
Multiplication and Division of ComplexNumbers in Polar Form
Let z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then
Product: z1z2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2 − sinθ1sinθ2) + i(cosθ1sinθ2 +sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.
N. B. Vyas Complex Numbers
![Page 39: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/39.jpg)
Multiplication and Division of ComplexNumbers in Polar Form
Let z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then
Product: z1z2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)
= r1r2[(cosθ1cosθ2 − sinθ1sinθ2) + i(cosθ1sinθ2 +sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.
N. B. Vyas Complex Numbers
![Page 40: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/40.jpg)
Multiplication and Division of ComplexNumbers in Polar Form
Let z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then
Product: z1z2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2 − sinθ1sinθ2) + i(cosθ1sinθ2 +sinθ1cosθ2)]
= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.
N. B. Vyas Complex Numbers
![Page 41: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/41.jpg)
Multiplication and Division of ComplexNumbers in Polar Form
Let z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then
Product: z1z2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2 − sinθ1sinθ2) + i(cosθ1sinθ2 +sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]
which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.
N. B. Vyas Complex Numbers
![Page 42: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/42.jpg)
Multiplication and Division of ComplexNumbers in Polar Form
Let z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then
Product: z1z2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2 − sinθ1sinθ2) + i(cosθ1sinθ2 +sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.
N. B. Vyas Complex Numbers
![Page 43: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/43.jpg)
Multiplication and Division of ComplexNumbers in Polar Form
Division:z1
z2=r1(cosθ1 + isinθ1)
r2(cosθ2 + isinθ2)
=r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)
r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)
=r1
r2[(cosθ1cosθ2 + sinθ1sinθ2) + i(sinθ1cosθ2 −
cosθ1sinθ2)]
=r1
r2[cos(θ1 − θ2) + isin(θ1 − θ2)]
which shows that the modulus and amplitude of thequotient of two complex numbers z1 and z2 is the
quotient of their moduli ( i.e.r1
r2) and difference of
their arguments( i.e. θ1 − θ2 ) respectively.
N. B. Vyas Complex Numbers
![Page 44: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/44.jpg)
Multiplication and Division of ComplexNumbers in Polar Form
Division:z1
z2=r1(cosθ1 + isinθ1)
r2(cosθ2 + isinθ2)
=r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)
r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)
=r1
r2[(cosθ1cosθ2 + sinθ1sinθ2) + i(sinθ1cosθ2 −
cosθ1sinθ2)]
=r1
r2[cos(θ1 − θ2) + isin(θ1 − θ2)]
which shows that the modulus and amplitude of thequotient of two complex numbers z1 and z2 is the
quotient of their moduli ( i.e.r1
r2) and difference of
their arguments( i.e. θ1 − θ2 ) respectively.
N. B. Vyas Complex Numbers
![Page 45: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/45.jpg)
Multiplication and Division of ComplexNumbers in Polar Form
Division:z1
z2=r1(cosθ1 + isinθ1)
r2(cosθ2 + isinθ2)
=r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)
r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)
=r1
r2[(cosθ1cosθ2 + sinθ1sinθ2) + i(sinθ1cosθ2 −
cosθ1sinθ2)]
=r1
r2[cos(θ1 − θ2) + isin(θ1 − θ2)]
which shows that the modulus and amplitude of thequotient of two complex numbers z1 and z2 is the
quotient of their moduli ( i.e.r1
r2) and difference of
their arguments( i.e. θ1 − θ2 ) respectively.
N. B. Vyas Complex Numbers
![Page 46: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/46.jpg)
Multiplication and Division of ComplexNumbers in Polar Form
Division:z1
z2=r1(cosθ1 + isinθ1)
r2(cosθ2 + isinθ2)
=r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)
r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)
=r1
r2[(cosθ1cosθ2 + sinθ1sinθ2) + i(sinθ1cosθ2 −
cosθ1sinθ2)]
=r1
r2[cos(θ1 − θ2) + isin(θ1 − θ2)]
which shows that the modulus and amplitude of thequotient of two complex numbers z1 and z2 is the
quotient of their moduli ( i.e.r1
r2) and difference of
their arguments( i.e. θ1 − θ2 ) respectively.
N. B. Vyas Complex Numbers
![Page 47: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/47.jpg)
Multiplication and Division of ComplexNumbers in Polar Form
Division:z1
z2=r1(cosθ1 + isinθ1)
r2(cosθ2 + isinθ2)
=r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)
r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)
=r1
r2[(cosθ1cosθ2 + sinθ1sinθ2) + i(sinθ1cosθ2 −
cosθ1sinθ2)]
=r1
r2[cos(θ1 − θ2) + isin(θ1 − θ2)]
which shows that the modulus and amplitude of thequotient of two complex numbers z1 and z2 is the
quotient of their moduli ( i.e.r1
r2) and difference of
their arguments( i.e. θ1 − θ2 ) respectively.N. B. Vyas Complex Numbers
![Page 48: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/48.jpg)
Exponential form of Complex Numbers
We know that
sinθ = θ − θ3
3!+θ5
5!− θ7
7!. . .
cosθ = 1− θ2
2!+θ4
4!− θ6
6!. . . and
eθ = 1 + θ +θ2
2!+θ3
3!+θ4
4!. . .
Now θ = iθ
eiθ = 1 + iθ +(iθ)2
2!+
(iθ)3
3!+
(iθ)4
4!. . .
eiθ = 1 + iθ − θ2
2!− iθ
3
3!+θ4
4!. . . {∵ i2 = −1, i3 = −i, i4 = 1
eiθ =
(1− θ2
2!+θ4
4!− . . .
)+ i
(θ − θ3
3!+θ5
5!− . . .
)eiθ = (cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 49: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/49.jpg)
Exponential form of Complex Numbers
We know that
sinθ = θ − θ3
3!+θ5
5!− θ7
7!. . .
cosθ = 1− θ2
2!+θ4
4!− θ6
6!. . . and
eθ = 1 + θ +θ2
2!+θ3
3!+θ4
4!. . .
Now θ = iθ
eiθ = 1 + iθ +(iθ)2
2!+
(iθ)3
3!+
(iθ)4
4!. . .
eiθ = 1 + iθ − θ2
2!− iθ
3
3!+θ4
4!. . . {∵ i2 = −1, i3 = −i, i4 = 1
eiθ =
(1− θ2
2!+θ4
4!− . . .
)+ i
(θ − θ3
3!+θ5
5!− . . .
)eiθ = (cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 50: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/50.jpg)
Exponential form of Complex Numbers
We know that
sinθ = θ − θ3
3!+θ5
5!− θ7
7!. . .
cosθ = 1− θ2
2!+θ4
4!− θ6
6!. . . and
eθ = 1 + θ +θ2
2!+θ3
3!+θ4
4!. . .
Now θ = iθ
eiθ = 1 + iθ +(iθ)2
2!+
(iθ)3
3!+
(iθ)4
4!. . .
eiθ = 1 + iθ − θ2
2!− iθ
3
3!+θ4
4!. . . {∵ i2 = −1, i3 = −i, i4 = 1
eiθ =
(1− θ2
2!+θ4
4!− . . .
)+ i
(θ − θ3
3!+θ5
5!− . . .
)eiθ = (cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 51: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/51.jpg)
Exponential form of Complex Numbers
We know that
sinθ = θ − θ3
3!+θ5
5!− θ7
7!. . .
cosθ = 1− θ2
2!+θ4
4!− θ6
6!. . . and
eθ = 1 + θ +θ2
2!+θ3
3!+θ4
4!. . .
Now θ = iθ
eiθ = 1 + iθ +(iθ)2
2!+
(iθ)3
3!+
(iθ)4
4!. . .
eiθ = 1 + iθ − θ2
2!− iθ
3
3!+θ4
4!. . . {∵ i2 = −1, i3 = −i, i4 = 1
eiθ =
(1− θ2
2!+θ4
4!− . . .
)+ i
(θ − θ3
3!+θ5
5!− . . .
)eiθ = (cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 52: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/52.jpg)
Exponential form of Complex Numbers
We know that
sinθ = θ − θ3
3!+θ5
5!− θ7
7!. . .
cosθ = 1− θ2
2!+θ4
4!− θ6
6!. . . and
eθ = 1 + θ +θ2
2!+θ3
3!+θ4
4!. . .
Now θ = iθ
eiθ = 1 + iθ +(iθ)2
2!+
(iθ)3
3!+
(iθ)4
4!. . .
eiθ = 1 + iθ − θ2
2!− iθ
3
3!+θ4
4!. . . {∵ i2 = −1, i3 = −i, i4 = 1
eiθ =
(1− θ2
2!+θ4
4!− . . .
)+ i
(θ − θ3
3!+θ5
5!− . . .
)eiθ = (cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 53: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/53.jpg)
Exponential form of Complex Numbers
We know that
sinθ = θ − θ3
3!+θ5
5!− θ7
7!. . .
cosθ = 1− θ2
2!+θ4
4!− θ6
6!. . . and
eθ = 1 + θ +θ2
2!+θ3
3!+θ4
4!. . .
Now θ = iθ
eiθ = 1 + iθ +(iθ)2
2!+
(iθ)3
3!+
(iθ)4
4!. . .
eiθ = 1 + iθ − θ2
2!− iθ
3
3!+θ4
4!. . . {∵ i2 = −1, i3 = −i, i4 = 1
eiθ =
(1− θ2
2!+θ4
4!− . . .
)+ i
(θ − θ3
3!+θ5
5!− . . .
)eiθ = (cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 54: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/54.jpg)
Exponential form of Complex Numbers
We know that
sinθ = θ − θ3
3!+θ5
5!− θ7
7!. . .
cosθ = 1− θ2
2!+θ4
4!− θ6
6!. . . and
eθ = 1 + θ +θ2
2!+θ3
3!+θ4
4!. . .
Now θ = iθ
eiθ = 1 + iθ +(iθ)2
2!+
(iθ)3
3!+
(iθ)4
4!. . .
eiθ = 1 + iθ − θ2
2!− iθ
3
3!+θ4
4!. . . {∵ i2 = −1, i3 = −i, i4 = 1
eiθ =
(1− θ2
2!+θ4
4!− . . .
)+ i
(θ − θ3
3!+θ5
5!− . . .
)
eiθ = (cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 55: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/55.jpg)
Exponential form of Complex Numbers
We know that
sinθ = θ − θ3
3!+θ5
5!− θ7
7!. . .
cosθ = 1− θ2
2!+θ4
4!− θ6
6!. . . and
eθ = 1 + θ +θ2
2!+θ3
3!+θ4
4!. . .
Now θ = iθ
eiθ = 1 + iθ +(iθ)2
2!+
(iθ)3
3!+
(iθ)4
4!. . .
eiθ = 1 + iθ − θ2
2!− iθ
3
3!+θ4
4!. . . {∵ i2 = −1, i3 = −i, i4 = 1
eiθ =
(1− θ2
2!+θ4
4!− . . .
)+ i
(θ − θ3
3!+θ5
5!− . . .
)eiθ = (cosθ + isinθ)
N. B. Vyas Complex Numbers
![Page 56: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/56.jpg)
Laws of Complex Numbers
If z1 and z2 are two complex numbers, then
1 Triangle Inequality: |z1 + z2| ≤ |z1|+ |z2|2 |z1 − z2| ≥ ||z1| − |z2||3 Parellelogram equality:|z1 + z2|2 + |z1 − z2|2 = 2(|z1|2 + |z2|2)
4 |z1z2| = |z1||z2|
5
∣∣∣∣z1
z2
∣∣∣∣ =|z1||z2|
N. B. Vyas Complex Numbers
![Page 57: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/57.jpg)
Laws of Complex Numbers
If z1 and z2 are two complex numbers, then
1 Triangle Inequality: |z1 + z2| ≤ |z1|+ |z2|
2 |z1 − z2| ≥ ||z1| − |z2||3 Parellelogram equality:|z1 + z2|2 + |z1 − z2|2 = 2(|z1|2 + |z2|2)
4 |z1z2| = |z1||z2|
5
∣∣∣∣z1
z2
∣∣∣∣ =|z1||z2|
N. B. Vyas Complex Numbers
![Page 58: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/58.jpg)
Laws of Complex Numbers
If z1 and z2 are two complex numbers, then
1 Triangle Inequality: |z1 + z2| ≤ |z1|+ |z2|2 |z1 − z2| ≥ ||z1| − |z2||
3 Parellelogram equality:|z1 + z2|2 + |z1 − z2|2 = 2(|z1|2 + |z2|2)
4 |z1z2| = |z1||z2|
5
∣∣∣∣z1
z2
∣∣∣∣ =|z1||z2|
N. B. Vyas Complex Numbers
![Page 59: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/59.jpg)
Laws of Complex Numbers
If z1 and z2 are two complex numbers, then
1 Triangle Inequality: |z1 + z2| ≤ |z1|+ |z2|2 |z1 − z2| ≥ ||z1| − |z2||3 Parellelogram equality:|z1 + z2|2 + |z1 − z2|2 = 2(|z1|2 + |z2|2)
4 |z1z2| = |z1||z2|
5
∣∣∣∣z1
z2
∣∣∣∣ =|z1||z2|
N. B. Vyas Complex Numbers
![Page 60: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/60.jpg)
Laws of Complex Numbers
If z1 and z2 are two complex numbers, then
1 Triangle Inequality: |z1 + z2| ≤ |z1|+ |z2|2 |z1 − z2| ≥ ||z1| − |z2||3 Parellelogram equality:|z1 + z2|2 + |z1 − z2|2 = 2(|z1|2 + |z2|2)
4 |z1z2| = |z1||z2|
5
∣∣∣∣z1
z2
∣∣∣∣ =|z1||z2|
N. B. Vyas Complex Numbers
![Page 61: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/61.jpg)
Laws of Complex Numbers
If z1 and z2 are two complex numbers, then
1 Triangle Inequality: |z1 + z2| ≤ |z1|+ |z2|2 |z1 − z2| ≥ ||z1| − |z2||3 Parellelogram equality:|z1 + z2|2 + |z1 − z2|2 = 2(|z1|2 + |z2|2)
4 |z1z2| = |z1||z2|
5
∣∣∣∣z1
z2
∣∣∣∣ =|z1||z2|
N. B. Vyas Complex Numbers
![Page 62: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/62.jpg)
Ex
1 Find complex conjugate of3 + 2i
1− i
N. B. Vyas Complex Numbers
![Page 63: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/63.jpg)
Theorem
DeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ
∴ z = x+ iy = r(cosθ + isinθ)zn = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)E.g.:
(cosθ + isinθ)2 = cos2θ + isin2θ
(cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ
N. B. Vyas Complex Numbers
![Page 64: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/64.jpg)
Theorem
DeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ
∴ z = x+ iy = r(cosθ + isinθ)
zn = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)E.g.:
(cosθ + isinθ)2 = cos2θ + isin2θ
(cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ
N. B. Vyas Complex Numbers
![Page 65: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/65.jpg)
Theorem
DeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ
∴ z = x+ iy = r(cosθ + isinθ)zn = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)E.g.:
(cosθ + isinθ)2 = cos2θ + isin2θ
(cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ
N. B. Vyas Complex Numbers
![Page 66: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/66.jpg)
Theorem
DeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ
∴ z = x+ iy = r(cosθ + isinθ)zn = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)
E.g.:
(cosθ + isinθ)2 = cos2θ + isin2θ
(cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ
N. B. Vyas Complex Numbers
![Page 67: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/67.jpg)
Theorem
DeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ
∴ z = x+ iy = r(cosθ + isinθ)zn = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)E.g.:
(cosθ + isinθ)2 = cos2θ + isin2θ
(cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ
N. B. Vyas Complex Numbers
![Page 68: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/68.jpg)
Theorem
DeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ
∴ z = x+ iy = r(cosθ + isinθ)zn = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)E.g.:
(cosθ + isinθ)2 = cos2θ + isin2θ
(cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ
N. B. Vyas Complex Numbers
![Page 69: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/69.jpg)
Theorem
DeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ
∴ z = x+ iy = r(cosθ + isinθ)zn = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)E.g.:
(cosθ + isinθ)2 = cos2θ + isin2θ
(cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ
N. B. Vyas Complex Numbers
![Page 70: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/70.jpg)
Examples
Ex. Simplify
(cos2θ + isin2θ)23(cosθ − isinθ)2
(cos3θ − isin3θ)2(cos5θ − isin5θ)13
N. B. Vyas Complex Numbers
![Page 71: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/71.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 72: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/72.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 73: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/73.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 74: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/74.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 75: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/75.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 )
and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 76: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/76.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 77: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/77.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 78: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/78.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 79: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/79.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 80: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/80.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 81: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/81.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 82: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/82.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 83: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/83.jpg)
Examples
Ex. Evaluate (1 + i√
3)90 + (1− i√
3)90
Sol. Let x = 1 and y =√
3
r =√x2 + y2 =
√1 + 3 =
√4 = 2
θ = tan−1(yx
)= tan−1
(√3
1
)=π
3
∴ 1 + i√
3 = 2(cosπ3 + isinπ3 ) and 1− i√
3 = 2(cosπ3 − isinπ3 )
∴ (1 + i√
3)90 + (1− i√
3)90
=[2(cosπ3 + isinπ3 )
]90+[2(cosπ3 − isin
π3 )]90
= 290(cos30π + isin30π) + 290(cos30π − isin30π)
= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
![Page 84: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/84.jpg)
Example
Ex. Evaluate
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n
N. B. Vyas Complex Numbers
![Page 85: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/85.jpg)
Sol. We have sin2θ + cos2θ = 1
∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)
∴1 + sinθ + icosθ
1 + sinθ − icosθ= sinθ + icosθ
= cos(π2 − θ
)+ isin
(π2 − θ
)∴
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n=[cos(π2 − θ
)+ isin
(π2 − θ
)]n= cos
[n(π2 − θ
)]+ isin
[n(π2 − θ
)]
N. B. Vyas Complex Numbers
![Page 86: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/86.jpg)
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)
∴1 + sinθ + icosθ
1 + sinθ − icosθ= sinθ + icosθ
= cos(π2 − θ
)+ isin
(π2 − θ
)∴
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n=[cos(π2 − θ
)+ isin
(π2 − θ
)]n= cos
[n(π2 − θ
)]+ isin
[n(π2 − θ
)]
N. B. Vyas Complex Numbers
![Page 87: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/87.jpg)
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)
∴1 + sinθ + icosθ
1 + sinθ − icosθ= sinθ + icosθ
= cos(π2 − θ
)+ isin
(π2 − θ
)∴
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n=[cos(π2 − θ
)+ isin
(π2 − θ
)]n= cos
[n(π2 − θ
)]+ isin
[n(π2 − θ
)]
N. B. Vyas Complex Numbers
![Page 88: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/88.jpg)
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ
= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)
∴1 + sinθ + icosθ
1 + sinθ − icosθ= sinθ + icosθ
= cos(π2 − θ
)+ isin
(π2 − θ
)∴
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n=[cos(π2 − θ
)+ isin
(π2 − θ
)]n= cos
[n(π2 − θ
)]+ isin
[n(π2 − θ
)]
N. B. Vyas Complex Numbers
![Page 89: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/89.jpg)
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)
= (sinθ + icosθ)(sinθ − icosθ + 1)
∴1 + sinθ + icosθ
1 + sinθ − icosθ= sinθ + icosθ
= cos(π2 − θ
)+ isin
(π2 − θ
)∴
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n=[cos(π2 − θ
)+ isin
(π2 − θ
)]n= cos
[n(π2 − θ
)]+ isin
[n(π2 − θ
)]
N. B. Vyas Complex Numbers
![Page 90: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/90.jpg)
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)
∴1 + sinθ + icosθ
1 + sinθ − icosθ= sinθ + icosθ
= cos(π2 − θ
)+ isin
(π2 − θ
)∴
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n=[cos(π2 − θ
)+ isin
(π2 − θ
)]n= cos
[n(π2 − θ
)]+ isin
[n(π2 − θ
)]
N. B. Vyas Complex Numbers
![Page 91: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/91.jpg)
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)
∴1 + sinθ + icosθ
1 + sinθ − icosθ= sinθ + icosθ
= cos(π2 − θ
)+ isin
(π2 − θ
)∴
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n=[cos(π2 − θ
)+ isin
(π2 − θ
)]n= cos
[n(π2 − θ
)]+ isin
[n(π2 − θ
)]
N. B. Vyas Complex Numbers
![Page 92: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/92.jpg)
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)
∴1 + sinθ + icosθ
1 + sinθ − icosθ= sinθ + icosθ
= cos(π2 − θ
)+ isin
(π2 − θ
)
∴
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n=[cos(π2 − θ
)+ isin
(π2 − θ
)]n= cos
[n(π2 − θ
)]+ isin
[n(π2 − θ
)]
N. B. Vyas Complex Numbers
![Page 93: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/93.jpg)
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)
∴1 + sinθ + icosθ
1 + sinθ − icosθ= sinθ + icosθ
= cos(π2 − θ
)+ isin
(π2 − θ
)∴
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n=[cos(π2 − θ
)+ isin
(π2 − θ
)]n
= cos[n(π2 − θ
)]+ isin
[n(π2 − θ
)]
N. B. Vyas Complex Numbers
![Page 94: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/94.jpg)
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)
∴1 + sinθ + icosθ
1 + sinθ − icosθ= sinθ + icosθ
= cos(π2 − θ
)+ isin
(π2 − θ
)∴
(1 + sinθ + icosθ
1 + sinθ − icosθ
)n=[cos(π2 − θ
)+ isin
(π2 − θ
)]n= cos
[n(π2 − θ
)]+ isin
[n(π2 − θ
)]
N. B. Vyas Complex Numbers
![Page 95: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/95.jpg)
Example
Ex. If z1 = eiθ1 , z2 = eiθ2 and z1 − z2 = 0 prove that1
z1− 1
z2= 0
N. B. Vyas Complex Numbers
![Page 96: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/96.jpg)
Example
Sol. Herez1 = eiθ1 = cosθ1 + isinθ1
z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0
Now1
z1− 1
z2=
1
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
=cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
![Page 97: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/97.jpg)
Example
Sol. Herez1 = eiθ1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2
Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0
Now1
z1− 1
z2=
1
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
=cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
![Page 98: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/98.jpg)
Example
Sol. Herez1 = eiθ1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2
Given that z1 − z2 = 0
(cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0
Now1
z1− 1
z2=
1
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
=cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
![Page 99: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/99.jpg)
Example
Sol. Herez1 = eiθ1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2
Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0
(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0
Now1
z1− 1
z2=
1
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
=cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
![Page 100: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/100.jpg)
Example
Sol. Herez1 = eiθ1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2
Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0
Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0
Now1
z1− 1
z2=
1
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
=cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
![Page 101: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/101.jpg)
Example
Sol. Herez1 = eiθ1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2
Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we get
cosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0
Now1
z1− 1
z2=
1
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
=cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
![Page 102: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/102.jpg)
Example
Sol. Herez1 = eiθ1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2
Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0
Now1
z1− 1
z2=
1
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
=cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
![Page 103: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/103.jpg)
Example
Sol. Herez1 = eiθ1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2
Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0
Now1
z1− 1
z2=
1
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
=cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
![Page 104: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/104.jpg)
Example
Sol. Herez1 = eiθ1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2
Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0
Now1
z1− 1
z2=
1
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
=cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
![Page 105: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/105.jpg)
Example
Sol. Herez1 = eiθ1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2
Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0
Now1
z1− 1
z2=
1
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
=cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
![Page 106: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/106.jpg)
Ex. (1 + i)n + (1− i)n = 2n2+1cosnπ4
N. B. Vyas Complex Numbers
![Page 107: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/107.jpg)
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
If n is any positive integer, then by De Moivre’s theorem(cos
θ
n+ isin
θ
n
)n= cos
(θ
nn
)+ isin
(θ
nn
)= cosθ + isinθ
Thus cosθ
n+ isin
θ
nis one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1n = cos
θ
n+ isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ
Hence the general form of complex number cosθ+ isinθ form restof all the roots.
N. B. Vyas Complex Numbers
![Page 108: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/108.jpg)
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
If n is any positive integer, then by De Moivre’s theorem
(cos
θ
n+ isin
θ
n
)n= cos
(θ
nn
)+ isin
(θ
nn
)= cosθ + isinθ
Thus cosθ
n+ isin
θ
nis one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1n = cos
θ
n+ isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ
Hence the general form of complex number cosθ+ isinθ form restof all the roots.
N. B. Vyas Complex Numbers
![Page 109: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/109.jpg)
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
If n is any positive integer, then by De Moivre’s theorem(cos
θ
n+ isin
θ
n
)n= cos
(θ
nn
)+ isin
(θ
nn
)= cosθ + isinθ
Thus cosθ
n+ isin
θ
nis one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1n = cos
θ
n+ isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ
Hence the general form of complex number cosθ+ isinθ form restof all the roots.
N. B. Vyas Complex Numbers
![Page 110: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/110.jpg)
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
If n is any positive integer, then by De Moivre’s theorem(cos
θ
n+ isin
θ
n
)n= cos
(θ
nn
)+ isin
(θ
nn
)= cosθ + isinθ
Thus cosθ
n+ isin
θ
nis one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1n = cos
θ
n+ isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ
Hence the general form of complex number cosθ+ isinθ form restof all the roots.
N. B. Vyas Complex Numbers
![Page 111: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/111.jpg)
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
If n is any positive integer, then by De Moivre’s theorem(cos
θ
n+ isin
θ
n
)n= cos
(θ
nn
)+ isin
(θ
nn
)= cosθ + isinθ
Thus cosθ
n+ isin
θ
nis one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1n = cos
θ
n+ isin
θ
n
The remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ
Hence the general form of complex number cosθ+ isinθ form restof all the roots.
N. B. Vyas Complex Numbers
![Page 112: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/112.jpg)
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
If n is any positive integer, then by De Moivre’s theorem(cos
θ
n+ isin
θ
n
)n= cos
(θ
nn
)+ isin
(θ
nn
)= cosθ + isinθ
Thus cosθ
n+ isin
θ
nis one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1n = cos
θ
n+ isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ
Hence the general form of complex number cosθ+ isinθ form restof all the roots.
N. B. Vyas Complex Numbers
![Page 113: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/113.jpg)
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
If n is any positive integer, then by De Moivre’s theorem(cos
θ
n+ isin
θ
n
)n= cos
(θ
nn
)+ isin
(θ
nn
)= cosθ + isinθ
Thus cosθ
n+ isin
θ
nis one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1n = cos
θ
n+ isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ
Hence the general form of complex number cosθ+ isinθ form restof all the roots.
N. B. Vyas Complex Numbers
![Page 114: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/114.jpg)
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
If n is any positive integer, then by De Moivre’s theorem(cos
θ
n+ isin
θ
n
)n= cos
(θ
nn
)+ isin
(θ
nn
)= cosθ + isinθ
Thus cosθ
n+ isin
θ
nis one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1n = cos
θ
n+ isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ
Hence the general form of complex number cosθ+ isinθ form restof all the roots.
N. B. Vyas Complex Numbers
![Page 115: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/115.jpg)
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
If n is any positive integer, then by De Moivre’s theorem(cos
θ
n+ isin
θ
n
)n= cos
(θ
nn
)+ isin
(θ
nn
)= cosθ + isinθ
Thus cosθ
n+ isin
θ
nis one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1n = cos
θ
n+ isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ
Hence the general form of complex number cosθ+ isinθ form restof all the roots.
N. B. Vyas Complex Numbers
![Page 116: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/116.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)which gives all the roots of (cosθ + isinθ)
1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
). . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers
![Page 117: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/117.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)
which gives all the roots of (cosθ + isinθ)1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
). . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers
![Page 118: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/118.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)which gives all the roots of (cosθ + isinθ)
1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
). . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers
![Page 119: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/119.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)which gives all the roots of (cosθ + isinθ)
1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
). . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers
![Page 120: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/120.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)which gives all the roots of (cosθ + isinθ)
1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
). . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers
![Page 121: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/121.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)which gives all the roots of (cosθ + isinθ)
1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)
k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
). . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers
![Page 122: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/122.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)which gives all the roots of (cosθ + isinθ)
1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
)
. . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers
![Page 123: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/123.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)which gives all the roots of (cosθ + isinθ)
1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
). . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers
![Page 124: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/124.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)which gives all the roots of (cosθ + isinθ)
1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
). . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers
![Page 125: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/125.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)which gives all the roots of (cosθ + isinθ)
1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
). . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]
The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers
![Page 126: Complex numbers 1](https://reader034.vdocument.in/reader034/viewer/2022050919/548060885906b51c298b4744/html5/thumbnails/126.jpg)
Root of a Complex Number
(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]
1n
= cos
(2kπ + θ
n
)+ isin
(2kπ + θ
n
)which gives all the roots of (cosθ + isinθ)
1n for
k = 0, 1, 2, . . . (n− 1).
This roots are as follows:
For k = 0, cosθ
n+ isin
θ
n
k = 1, cos
(2π + θ
n
)+ isin
(2π + θ
n
)k = 2, cos
(4π + θ
n
)+ isin
(4π + θ
n
). . . . . .
. . . . . .
k = n− 1, cos
[2(n− 1)π + θ
n
]+ isin
[2(n− 1)π + θ
n
]The further values of k will give the same roots as above in order.
N. B. Vyas Complex Numbers