consider the quadratic equation x 2 + 1 = 0. solving for x , gives x 2 = – 1
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Complex Numbers. Consider the quadratic equation x 2 + 1 = 0. Solving for x , gives x 2 = – 1. We make the following definition:. Complex Numbers. Note that squaring both sides yields: therefore and so and. And so on…. - PowerPoint PPT PresentationTRANSCRIPT
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Consider the quadratic equation x2 + 1 = 0. Solving for x , gives x2 = – 1
12 x
1x
We make the following definition:
1i
Complex Numbers
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1i
Complex Numbers
12 iNote that squaring both sides yields:therefore
and
so
and
iiiii *1* 13 2
1)1(*)1(* 224 iii
iiiii *1*45
1*1* 2246 iiii
And so on…
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Real NumbersImaginary Numbers
Real numbers and imaginary numbers are subsets of the set of complex numbers.
Complex Numbers
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Definition of a Complex Number Definition of a Complex Number
If a and b are real numbers, the number a + bi is a complex number, and it is said to be written in standard form.
If b = 0, the number a + bi = a is a real number.
If a = 0, the number a + bi is called an imaginary number.
Write the complex number in standard form
81 81 i 241 i 221 i
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Addition and Subtraction of Complex Addition and Subtraction of Complex Numbers Numbers
If a + bi and c +di are two complex numbers written in standard form, their sum and difference are defined as follows.
i)db()ca()dic()bia(
i)db()ca()dic()bia(
Sum:
Difference:
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Perform the subtraction and write the answer in standard form.
( 3 + 2i ) – ( 6 + 13i ) 3 + 2i – 6 – 13i –3 – 11i
234188 i
234298 ii
234238 ii
4
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Multiplying Complex NumbersMultiplying Complex Numbers
Multiplying complex numbers is similar to multiplying polynomials and combining like terms.
Perform the operation and write the result in standard form. ( 6 – 2i )( 2 – 3i )
F O I L12 – 18i – 4i + 6i2
12 – 22i + 6 ( -1 )6 – 22i
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Consider ( 3 + 2i )( 3 – 2i )9 – 6i + 6i – 4i2
9 – 4( -1 )9 + 4 13
This is a real number. The product of two complex numbers can be a real number.
This concept can be used to divide complex numbers.
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Complex Conjugates and Division Complex Conjugates and Division
Complex conjugates-a pair of complex numbers of the form a + bi and a – bi where a and b are real numbers.
( a + bi )( a – bi )a 2 – abi + abi – b 2 i 2
a 2 – b 2( -1 )a 2 + b 2
The product of a complex conjugate pair is a positive real number.
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To find the quotient of two complex numbers multiply the numerator and denominator by the conjugate of the denominator.
dic
bia
dic
dic
dic
bia
22
2
dc
bdibciadiac
22 dc
iadbcbdac
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Perform the operation and write the result in
standard form.
i
i
21
76
i
i
i
i
21
21
21
76
22
2
21
147126
iii
41
5146
i
5
520 i
5
5
5
20 i i4
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ii
i
4
31 i
i
ii
i
i
i
4
4
4
31
Perform the operation and write the result in standard form.
222
2
14
312
i
i
ii116
312
1
1
ii
ii17
3
17
121 ii
17
3
17
121
i17
317
17
1217
i
17
14
17
5
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Expressing Complex NumbersExpressing Complex Numbers in Polar Form in Polar Form
Now, any Complex Number can be expressed as:X + Y i That number can be plotted as on ordered pair in rectangular form like so…
6
4
2
-2
-4
-6
-5 5
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Expressing Complex NumbersExpressing Complex Numbers in Polar Form in Polar Form
Remember these relationships between polar and rectangular form:
x
ytan 222 ryx
cosrx sinry
So any complex number, X + Yi, can be written inpolar form: irrYiX sincos
)sin(cossincos irirr
rcisHere is the shorthand way of writing polar form:
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Expressing Complex NumbersExpressing Complex Numbers in Polar Form in Polar Form
Rewrite the following complex number in polar form: 4 - 2i
Rewrite the following complex number inrectangular form: 0307cis
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Expressing Complex NumbersExpressing Complex Numbers in Polar Form in Polar Form
Express the following complex number inrectangular form: )
3sin
3(cos2
i
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Expressing Complex NumbersExpressing Complex Numbers in Polar Form in Polar Form
Express the following complex number inpolar form: 5i
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Products and Quotients of Products and Quotients of Complex Numbers in Polar FormComplex Numbers in Polar Form
)sin(cos 111 ir
The product of two complex numbers, and
Can be obtained by using the following formula:)sin(cos 222 ir
)sin(cos*)sin(cos 222111 irir
)]sin()[cos(* 212121 irr
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Products and Quotients of Products and Quotients of Complex Numbers in Polar FormComplex Numbers in Polar Form
)sin(cos 111 ir
The quotient of two complex numbers, and
Can be obtained by using the following formula:)sin(cos 222 ir
)sin(cos/)sin(cos 222111 irir
)]sin()[cos(/ 212121 irr
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Products and Quotients of Products and Quotients of Complex Numbers in Polar FormComplex Numbers in Polar Form
Find the product of 5cis30 and –2cis120
Next, write that product in rectangular form
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Products and Quotients of Products and Quotients of Complex Numbers in Polar FormComplex Numbers in Polar Form
Find the quotient of 36cis300 divided by 4cis120
Next, write that quotient in rectangular form
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Products and Quotients of Products and Quotients of Complex Numbers in Polar FormComplex Numbers in Polar Form
Find the result ofLeave your answer in polar form.
Based on how you answered this problem, what generalization can we make aboutraising a complex number in polar form toa given power?
4))120sin120(cos5( i
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De Moivre’s TheoremDe Moivre’s Theorem
De Moivre's Theorem is the theorem which shows us how to take complex numbers to any power easily.
De Moivre's Theorem – Let r(cos +isin ) be a complex number and n be any real number. Then
[r(cos +isin ]n = rn(cos n+isin n)
What is this saying?
The resulting r value will be r to the nth power and the resulting angle will be n times the original angle.
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De Moivre’s TheoremDe Moivre’s Theorem
Try a sample problem: What is [3(cos 45+isin45)]5?
To do this take 3 to the 5th power, then multiply 45 times 5 and plug back into trigonometric form.
35 = 243 and 45 * 5 =225 so the result is 243(cos 225+isin 225)
Remember to save space you can write it in compact form.
243(cos 225+isin 225)=243cis 225
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De Moivre’s TheoremDe Moivre’s Theorem
Find the result of: Because of the power involved, it would easier to change this complex number into polar form and then use De Moivre’s Theorem.
4)1( i
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De Moivre’s TheoremDe Moivre’s Theorem
De Moivre's Theorem also works not only for integer values of powers, but also rational values (so we can determine roots of complex numbers).
pp rcisyix11
)()(
)()1
*(11
pcisr
pcisr pp
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De Moivre’s TheoremDe Moivre’s Theorem
Simplify the following: 3
1
)344( i
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De Moivre’s TheoremDe Moivre’s Theorem
Every complex number has ‘p’ distinct ‘pth’ complex roots (2 square roots, 3 cube roots, etc.)
To find the p distinct pth roots of a complex number, we use the following form of De Moivre’s Theorem
)360
()(11
p
ncisryix pp
…where ‘n’ is all integer values between 0 and p-1. Why the 360? Well, if we were to graph the complexroots on a polar graph, we would see that the p rootswould be evenly spaced about 360 degrees (360/p wouldtell us how far apart the roots would be).
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De Moivre’s TheoremDe Moivre’s Theorem
Find the 4 distinct 4th roots of -3 - 3i
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De Moivre’s TheoremDe Moivre’s Theorem
Solve the following equation for all complexnumber solutions (roots):
0273 x