(acmsm077)mathsbooks.net/nelson specialist year 12/by chapter/ch 2...chapter 2: complex numbers and...

CARTESIAN FORMS

• review real and imaginary parts Re(z) and Im(z) of a complex number z (ACMSM077)• review Cartesian form (ACMSM078) • review complex arithmetic using Cartesian forms. (ACMSM079)

COMPLEX ARITHMETIC USING POLAR FORM

• use the modulus lzl of a complex number z and the argument Arg (z) of a non-zero complex number z andprove basic identities involving modulus and argument (ACMSM0S0)

• convert between Cartesian and polar form (ACMSM0SI)• define and use multiplication, division, and powers of complex numbers in polar form and the geometric

interpretation of these (ACMSM082)• prove and use De Moivre's theorem for integral powers. (ACMSM083)@

f'J111 REVIEW OF COMPLEX

NUMBERS

In this chapter you will review the concepts you studied in Year 11 relating to Complex Numbersand then develop further concepts. Recall the following definitions and rules.

IMPORTANT

The imaginary number i is the number such that i = �-

A complex number is a number that can be written in the form a + ib, where a and b are realnumbers.A complex number is often denoted by the letter z, so z =a+ ib.

0 Example 1

Solve the quadratic equation x2 - 8x + 25 = 0, expressing your answers in the form a+ bi, where

a, b ER.

Solution

Note that x2 - 8x + 25 does not factorise, so it

is necessary to solve by using the quadraticformula (or by completing the square).

Substitute and simplify.

Recall that H6 = �36 x (-1)=6Xi

76 I NELSON SENIOR MATHS Specialist 12

-b±�b2 -4ac x=-----2a

-(-8)±�(-8)2 -4(1)(25)X = ----'---------

2(1)8±H6=---

2

8±6i :.x=--

=4±3i

9780170250306

Alternate solution Complete the square.

Make a difference of two squares by first . . 9 9·2 wntmg+ =- 1.

Isolate x.

Recall the definition of the complex conjugate.

x2 - 8x + 16 + 9 = 0

(x-4)2 + 9 = 0

(x-4)2-9i2 =0 (x-4-3i)(x-4+ 3i) = 0

:. x= 4± 3i.

IMPORTANT

For a complex number z, where z =a+ ib (where a and bare real numbers), the complex

conjugate is z = a - ib.

0 Example 2

If z = x + yi, where x, y E R, prove that z z is always real.

Solution

Note that z = x-yi.

Expand and simplify.

There are no terms that include an i, so all

terms are real.

zz = (x+ yi)(x-yi) - 2 2,2 zz=x -y 1

=x2 -[y2 x (-1)] =x2

+ y2

Since zz = x2 + y2, zz is always real.

Recall that it is possible to represent the complex number z = x + yi (where x, y E R) geometrically

on an Argand diagram or Argand plane.

IMPORTANT

The complex number z = x + yi ( where x, y E R) can be represented geometrically on an

Argand diagram as the point P(x, y) or the vector z or OP.

9780170250306

y

P(x,y)

Y ------------r

0

I

I

I

I

I

X

Im (z)

P(z)

X Re (z)

CHAPTER 2: Complex numbers and De Moivre's theorem I 77

0 Example 3

For the complex number z = 3 - 2i, a state the point P representing z on an Argand plane b plot the vectors z and z that represent z and its conjugate z on an Argand plane c describe the relationship between the vectors z and z.

Solution

a The point P representing z is an ordered pair.

b The conjugate z is 3 + 2i.

P(3, -2).

y

2

0

-1

-2

2 3 4

t -- _L _ __ -

I I l -- --

X

c The vectors z and z have the same real part and opposite imaginary parts.

The conjugate vectors are reflections of each other over the real axis.

EXERCISE 2.01 Review of complex numbers

Concepts and techniques

$j.jnHM Use the quadratic formula to solve each equation, giving your solutions in the form a + bi, where a, b E R. a x2+2x+3=0C z2-5iz-6=0

b x2 -4x+7=0 d w2+w+2=0

2 Solve each equation below by completing the square. Check that your solutions are in the form a ± bi, where a, b E R. a x2-2x+ 2 = 0 C z2 -6z + 13 = 0

b y2+4y+5=0 d w2+8w+ 18=0.

3 Evaluate each of the following, giving your answer in simplest form. c i

so

78 I NELSON SENIOR MATHS Specialist 12 9780170250306

4 4;.;,,;.jfj Consider the complex number w = 3 - Si. a State the conjugate w. b Find the value of w x w.

5 Write a quadratic equation in the form az!- + bz + c = 0 with the following roots. a 2 ± i b -J3 ±2i C -l±i.Js

6 Using your answers to question 2, complete the sentence. If a quadratic equation has real coefficients and one root is p + qi, where p, q E R, then the other root will be ....

7 •d·iui•iM The vectors u, v and w represent the complex numbers u = I - 2i, v = 2 + i and w = -3 - 3i. Sketch u, v and won an Argand diagram.

Reasoning and communication

8 By considering the coefficients of the equation x2 + 2x + 5 = 0, decide whether or not the roots are complex conjugates. Solve the equation to see if your prediction is true.

9 Show that the roots of x2 - 3ix - 3 + i = 0 are 1 + i and 2i - 1. Explain why the roots are not

complex conjugates. r=l97

1 0 Evaluate I, i' . r=O

11 Consider the point P representing the complex number z = -2 + 3i. a Sketch the vector OP on an Arga_nd plane. b Sketch the vector OQ representing w = -2 + 3i . c Describe the geometric relationship between OP and OQ.

d Explain why OP - OQ lies on the y-axis.

tl•t..J REVIEW OF COMPLEX NUMBER

OPERATIONS

You can perform all four operations with complex numbers: addition, subtraction, multiplication and division. This relies on grouping or equating the real and imaginary parts.

IMPORTANT

For complex numbers a+ ib and c + id (where a, b, c and dare real numbers), a+ ib = c + id if and only if a = c AND b = d.

9780170250306 CHAPTER 2: Complex numbers and De Moivre's theorem I 79

0 Example 4

If z = 2 + i and w = -3 + 2i, find a 2z-w b w2

Solution

a Substitute and group the real and imaginary parts.

b Substitute and expand using the identity (a+ b)

2 = a

2 + 2ab + b2•

Simplify and recall that i2 = -1.

2z-w= 2(2+ i)- (-3 +2i) =4+2i+ 3 -2i =7

w2 = (-3 + 2i)2

= (-3)2 + 2(-3)(2i) + (2i)2

= 9- 12i + 4i2

= 9-12i-4 = 5-12i

Remember that you multiply by the complex conjugate to realise the denominator.

IMPORTANT

To realise the denominator of a complex number, multiply the number by 1 in the form�z

0 Example 5

Express 3 + i in the form a + bi, where a, b are real.3-i

Solution

Realise the denominator by multiplying by the conjugate.

Simplify and split into real and imaginary parts.

Write the answer in its simplest form by cancelling any common factors.


3+i 3+i 3+i-=-X-

3-i 3-i 3+i

9+6i+i2

=---

9-i2

9+6i-1 =---

9-(-1) 8+6i

=

10 8 6 .

=-+-t 10 10 3+i 4 3. -=-+-1

3-i 5 5

9780170250306

Recall the notation and language used for the real and imaginary parts of a complex number.

IMPORTANT

The real part of z =a+ ib is denoted by Re (z), where Re (z) = a and the imaginary part by Im(z), where Im (z) = b.If Re (z) = 0, then z is purely imaginary.Iflm (z) = 0, then z is purely real or just real .

0 Example 6

If z = 5 + 7i-i(4- 6i), find a Re (z)

Solution

b Im (z)

First expand and group into the real and imaginary parts.

a If z = a + bi, where a, b E R, then Re (z) = a

b If z = a + bi, where a, b E R, then Im (z) = b

z=5+7i-i(4-6i)= 5 + 7i - 4i + 6i2

= 5 + 7i-4i-6=-1 + 3i

Re (z) =-1

Im (z) = 3

EXERCISE 2.02 Review of complex number operations


•%\•111H4• Simplify each of the following expressions. a 4(2 + i) + 3(2 - i) b 3(5 - 4i) - i(3 - 2i)d (.J2+3i)(.J2-3i) e (9-i)(5+2i)

2 Find the values of u and v, where u and v are real, if

C (7 + 3i)2

a u +vi= 2(3 - 4i) + i(5 + i) b u +vi= -4(1 + 2i) - 3i(2 - 4i) 3 If z = 2 - i and w = -4 + Si, find:

a zw b z-w- 2

C Z XW d (z+ i)(w- 1) 4 •iHnHiP Realise the denominator for each of the following complex fractions.

a _l_ b l+ 2i C .js - i

2 - i 3 - 2i .Js + i5 Show that 1 = x - y - i ( x + y).

x-y+i(x+y) 2(x 2+y2 )

Complex number operations


l+i..J3 l-i../3 6 Show that�+� is always real.l-iv3 l+iv3 7 ;¥1-i,,j.ji4 Find Re (z) and Im (z) for each of the following.

a z=-2..J3-3iJib z = 2(5 + 6i) + 3(1 -7i) c z = x -yi -4w + vi, where x, y, w and v are real.

3-iJi dz=---l+iJiReasoning and communication

x -1+ yi 8 Find Re (z) and Im (z) if z = . , where x, y are real.

x-l-y1 9 Given z = u + vi, where u, v are real, find=· Hence show that

z

Re(_:.)= u2 -v2 and Im(_:.)=�z u2

+ v2 z u2

+ v2

10 S. l'fy 1 1Imp I

2 + 2. (x-yi) (x+ yi)

tl•JJ COMPLEX NUMBERS IN

POLAR FORM

You know that a complex number z can be expressed in the form x + yi, where x, y E R.This form is known as the Cartesian form (or rectangular form) as it uses x and y coordinates thatcan refer to axes. It is also convenient to express a complex number z in polar form, which uses thesize of z and the angle that z makes with the positive x-axis.

The argument of the complex number z -:f. 0, arg (z), isthe angle 0 that OP makes with the positive real axis,where P is the point that represents z in the complexplane. The principal value of the argument is the onein the interval (-7t, 7t]. The argument of O is not defined.The modulus of z is the magnitude of the vector z,

given by mod (z) = Jzl = �x2 + y2 .The polar form of z is given by z = r[cos (0) + i sin (0)],cos (0) + i sin (0) is often abbreviated to cis (0).The polar form is sometimes referred to as themodulus-argument form.


Im (z)

0

IMPORTANT

p

y = rsin (0)

0

x = rcos (0) Re (z)

9780170250306

0 Example 7

Write down the complex numbers in polar form, given the following arguments and moduli. a arg (z) = �, mod (z) = 63 b arg (w)=3:,lwl=2Solution

a Recall that arg (z) = 0, mod (z) = r.Substitute these into the formula.

b Recall that lwl = r. Substitute in the values.

z = r [ cos ( 0) + i sin ( 0)] = 6 [ cos(-j )+ i sin( -j)]

w = r [ cos(0) + i sin(0)] = 2 [ cos ( 3t ) + i sin ( 3t ) ]

IMPORTANT

The Cartesian and polar forms of the complex number z are related by the following equations. r = � x2 + y2 , tan (0) = 2'.., x = r cos (0) and y = r sin(0).

X

You can convert complex numbers between the Cartesian and polar forms using the relationships above.

0 Example 8

Convert the complex numbers to polar form. a z=l+i-.J3 b u=-2-2i

Solution

a Find r.

Find 0.

Write in polar form.

b Find r.

Find the principle value of 0.

Write in polar form.

zl =r=�l2 +(-./3)2 = 2 1 . ) .Ji 7t

COS (0) = -> 0, Sill (0 =-> 0 => 0 =-2 2 3

z= r [ cos(0)+ isin(0)] = 2 [ cos G) + j sin ( �) J

-2 -1 cos ( 0) = � = ,:- < 02'/2 '</2

-2 -1 37t sin(0)= �= ,:- <0 =>0=--2'</2 '</2 4

u=r[ cos(0)+ isin(0)] = 24[ cos (-:n) + isin (-:n )]

WS

Modulus and argument


ws

Complex number conver�ion�

0 Example 9

Convert each complex number below to Cartesian form. a z = 4[ cos(2f )+isin( 2;)] b v = h[ cos(¾) -i sin(¾)]Solution

a Evaluate cos ( 2f) and sin ( 2f) .

Expand. =-2+ 2i-J3

b Evaluate cos ( ¾) and sin ( ¾) . Expand. =1-i

Note that a complex number in the form z = r[ cos (0) - i sin (0)] can be converted to polar form using trigonometric identities. You can write z = r[ cos (0) - i sin (0)] = r[ cos (-0) + i sin (-0)] because cos (-0) = cos (0) and sin (-0) = -sin (0).

EXERCISE 2.03 Complex numbers in polar form


ii•inHd Find the complex number in the form r[cos (0) + i sin (0)] for which a arg(z)=1t,lzl=S b arg(z)=¾,mod(z)=4 C 0 = -t , r = 2 d arg ( V) = -71t , mod ( V) = 3h

2 For each complex number below, state its argument and modulus. a 2[ cos(¾)+isin(¾)] b 2.J2[ cos( 5;)+isin( 5t)] c cos(-91t )+ i sin(-91t) d � [ cos(-37t)+i sin(-r)]

3 ;;.;,;;.;o For each complex number z below, calculate mod (z) and arg (z) in exact form. a z=-J3+i d z=-�+ ,fi i

2 2

b z=3+ 3i e z=7i

1 1 . C z = ---t

2 2

z=-6 4 Convert each complex number below to polar form.

Ji 1 . a z=h+ih b w=-l+i-J3 d 1 1. .r;:5 v = ---:Ji---:Jit e z = -z-.p

C U=---1

2 2

w= 1

5 ii-inHP Convert each complex number below to Cartesian form. a 12[ cos(�)+isin(�)]

c cosG)+isinG)

b .J2 [ cos ( 3t ) + i sin ( 3t)]

cos(-�n) + i sin(-�n)d 2


6 Express each complex number below, with given arg and mod values, in Cartesian form. a arg(z)=- ¾, lzl = 8 b arg(z)= 5f, mod (z) = 3

. C 0=7t,r=9 d arg(v)=J, mod(v)=-J27 7 Simplify each complex number below, expressing your answer in x + yi form.

a 2cos(¾)+3isin(J) b .Js[ cos(-�n )+isin(¾)]


8 Use trigonometric identities to convert each complex number to polar form. a r[cos (0) - i sin (0)) b r[-cos (0) + i sin (0)) c -r[cos (0) + i sin (0)]

9 a Show that the points given by P(r cos (0), r sin (0)) for O � 0 < 27t form a circle in the Cartesian plane.

b Show that the complex number r sin (0) + ri cos (0) can be expressed in polar form. 2 2

c If x + yi = 3 cos (0) + 4i sin (0), show that �+L = 1. 9 16

tl1J1 MODULUS, ARGUMENT AND PRINCIPAL VALUE

Sn and_.::. represent the same angle on the Cartesian or Argand plane. This means that a complex3 3 . - . -number can also be expressed in infinitely many ways. For example, z = 2 [ cos ( 3:) + i sin ( 3:)] is the same as z = 2[ cos(-!n ) + i sin(-!n)] . To avoid ambiguity, a complex number is usually expressed in polar form using the principal argument. In this case, z = 2[ cos(3;) + i sin( 3:)] rather than z = 2[ cos(-�n ) + i sin(-�n )] is the convention since -7t < 3: � 7t and 3: is the principal argument.Multiplying and dividing complex numbers in modulus-argument form is much easier than in the Cartesian form. Adding and subtracting complex numbers is easier using Cartesian form.

0 Example 10

Evaluate z1 x z2 if z1 = 2[ cos(¾)+ i sin(¾)] and z2 = s[ cos(¾)+ i sin(¾)].

Solution

Use the distributive law. Rearrange into real and imaginary parts. Use the trigonometric identities.

9780170250306

z1 x z

2 = 2[ cos(¾ )+isin( ¾ )]x s[ cos(¾)+i sin(¾)]

= 10[ cos(¾ )cos(¾)+ icos(¾ )sin(¾)+ icos(¾ )sin(¾)+ i2 sin(¾ )sin(¾)]

= 10{ cos(¾ )cos(¾)-sin(¾ )sin(¾)+ i[ cos(¾ )sin(¾)+ cos(¾ )sin(¾)]}

=10[ cos(¾+¾)+isin(¾+¾)] .'.z1 xz2 =10[cos(�1)+isin(�1)]


The example above demonstrates that when multiplying, you multiply two complex numbers in polar form by multiplying the moduli and adding the arguments. IMPORTANT

If z1 = ri [cos (0 1 ) + i sin (0 1)1 and z2 = r2

[cos (02) + i sin (02)1 are two complex numbers, then • their product is z 1 z2 = r

1r

2[cos (0 1 + 02) + i sin (0 1 + 02)1

• arg (z1 z2) = arg (z 1) + arg (z2) and mod (z 1 z2) = mod (z 1 ) x mod (z2)

0 Example 11

Find the product of z 1 = 3[ cos(3;)+isin(3;)] and z2

= 2[ cos( 4;)+ i sin( 45n)]. Solution

Use the rule z 1 z2 = r1r

2[cos (0 1 + 02) + isin (0 1 + 02)1

z 1 z2 = 3[ cos( 3; )+ i sin( 35n)] x 2[ cos( 4t) + i sin( 4t)]= 3 x 2 [ cos ( 3; + 4t ) + i sin ( 3; + 45n ) ]

Simplify. Note that the principal value of the argument must be in the interval (-1t, 1tl. z1 z2 =6[ cos(7;)+isin(75n)]

= 6[ cos(-�n ) + i sin(-�n )]In a similar way, it is possible to develop rules for division of two complex numbers in polar form.

IMPORTANT

If z 1 = 1\ [ cos (0 1) + i sin (0 1) l and z2 = r2

[ cos (02) + i sin (02) l are two complex numbers, then �=l [cos(01

-02)+isin(01

- 0i )J. Z2 r2 If z = r[ cos (0) + i sin (0) l is a complex number, then

OR

z-1 = .!. = .!.[ cos(-0)+i sin(-0)].z r

If z 1 = ri[cos (0 1) + i sin (0 1)1 and z

2 = r

2[cos (02) + i sin (02)1 are two complex numbers, then

arg(�J=arg(z1 )- arg(zi ) and mod(�J= modt1�. Z2 Z2 mod Z2 If z = r[ cos (0) + i sin (0) l is a complex number, then arg(z-1)=-arg(z) and mod(z-1 )= mo�(z)"


0 Example 12

If z = Fi[ cos(¾)+ isin(¾)] and w = 2'1'2[ cos(2;)+ isin(23n) ], find

Solution

a Use the rule .£= Ji [cos(�- 2n )+isin(�- 2n)]IV 2/2 6 3 6 3

1= l[cos(01 -02

)+isin(01 -02 )],

Z2 r2

then simplify. 1 [ ( 37!) . . ( 37! )]=2 cos -6 +zsm -6

= ½[ cos(-;)+i sin(-;)]

b Use the rule z-'=½=Hcos(-0)+isin(-0)] -I 1 [ ( 7t) . . ( 7t )]z = Ji cos -6 +ism -6

EXERCISE 2.04 Modulus, argument and principal value

Concepts and te<:hniques

• • · • Expand, then use the identities sin (A+ B) = sin (A) cos (B) + cos (A) sin (B) andcos (A + B) = cos (A) cos (B) - sin (A) sin (B) to simplify: a [ cos(i)+isin(i) ][ cos(;)+ism(;)]

b. [ cos( 29n )+ isin( �n ) ][ cos( i )+ i sin( i)]c [ cos ( a.) + i sin ( a.)][ cos ( 4a.) + i sin ( 4a.)) d [cos (-3�) + i sin (-3�)][cos (-7�) + i sin (- 7�))

2 Express each of the following using a principal argument. a cis(3;) b cis( 5:) C 3 cis( 9t) d 2 cis(-�n )

3 Expand and then simplify, expressing your answer in polar form. [ cos(i)-ism(i) ][ cos( 3;)-isin(3;)]

4 -¥1H::i:iiJ11 Use the fact that for two complex numbers, z1z

2 = r1cis (0 1 ) x r2cis (02)

= r1r2cis (0 1 + 0

2) to simplify the following, leaving your answer in polar form.

a Fl cis(i)x)scis(¾) b 2 cis(5;)x4 cis(3;) c -3 cisrn-) x cis( ¾) d --/6 cis(-�n) x Fl cis(-�n)

5 Realise the denominators of the following, expressing your answer in polar form. 1 1 1

a -��-�� b -��-�� C cos( "I )-i sin( "I) sin(3; )+ icos(3;) 2 cis ( -:)


r.••••n z1 r1cis(01) r1 ( ) 1 1 1 1 ( )6 •n1m·1(ilff Use the rules - = . -cis 01 -02

and z- = -= -. -= -cis -0 t . l'fy z

2 r2cis(0

2 ) r2 z r c1s(0) r o s1mp 1 : 6 cis(½)

a 2 cis(-j) 12 cis (-31t)b 3 cis (2f)

d 1 e 1cis(0) 3 cis(3f)

7 If z = .J3cis(1") and w = cis( 3:), find:a zw b z

w

e (zwr2 zw

C

1 zw

d z2

8 Simplify each of the following, leaving your answer in polar form. a [ cos(-j )+i sin(-j) ][ cos( s; )-i sin( s;)]b [cos (4) - i sin (4)][cos (-2) - i sin (-2)] c 3[ cos(� ) -isin( �) ][ 4cos(3: )+ 4i sin( 3:)]


9 If z = cos (0) + i sin (0), prove that z-1 = z.10 If z = r[cos (0) + i sin (0)], prove that z-1 = �.

r

lScis(-t) C 5 cis(-�n)

4 cis(-!n)

11 If z = r[ cos (0) + i sin (0)] and w = r[cos (ex)+ i sin (ex)], show that Re (zw) = r2cos (0 + ex) and

Im (zw) = r2sin (0 + ex).

tl•J.1 OPERATIONS IN POLAR FORM

You can combine rules to solve problems.

0 Example 13

By first expressing .J3 + i in polar form, find: a arg( .J3 +i( b I( .J3 +i( I Solution

Find the modulus, r, and argument, 0, of .J3+i.


r=IF3+il=�(F3)2 +12 =2 J; , )1 1t cos(0) = 2, sm(0 = 2 => 0 = 6

: . .J3 +i = 2[ cos(¾)+isin(¾)]

9780170250306

a Use the rule arg (z-1) = -arg (z).

b Use the rule mod( z-1 ) = l ( )' mod z

7t = l(F3+ifl = mod(�+i)

= f

You can extend rules to multiple complex numbers. IMPORTANT

If z1 = r

1[cos (0

1) + i sin (01 )], z2

= r2[cos (02) + i sin (0

2)], ... , z

,, = r,,

[cos (0,,) + i sin (011)] are multiple complex numbers, then

z 1z2 . . . z,, = r

1r2 ... r11 [ cos (0

1 + 02 + ... + 011

) + i sin (01 + 02 + ... + 011)]

OR

If z1 = z

2 = ... = z,,, then the following rules also hold.

IMPORTANT

arg (z1') = n x arg (z) and lz1'1 = izl"and

0 Example 14

If z = .Jlcis(3;), w = F3cis( ¾) and u = .Jlcis( T), finda arg (zwu) b lw�J Solution

a Use the rule arg (z1z

2 . . . z

11) = arg (z1) + arg (z2) + ... + arg (z11)

Express 3; in the domain (-7t,7t]. b Use lz 1Z2

.. · znl = iziilz2 I .. · 12111 and lz1'1 = izl"Now substitute and simplify.

9780170250306

37t 7t 7t 37t arg( ZWU) = -+-+- = -4 4 2 2

-7t :. arg(zwu) = -. 2


Constructing spirals

w

Polar complex number operations

A number of difficult calculations can be done by expressing complex numbers in polar form. 0 Example 15

Evaluate ( r;;)' leaving your answer in polar form.(l+i) l-i-v3

Solution

First convert 1 + i and 1-i-J3 to polarform.

Now use the rules arg (zw) = arg (z) +arg (w) and arg (z-1) = -arg (z)

Use the rules lzwl = izllwl and iz-"I = izl-11-

Express your answer in the form r[cos (0) + isin (0)].

l +i =-12[ cos( �)+isin(�)] andl-i../3 = 2[ cos(-3n)+isin(-3n) Jarg[ . (

1 ·J3)]=-arg[(l+i)(l-i'-8)](1+1) l-1 3 =-[ arg(l+i)+arg(l-i'-8)] =-(�-�)

12

1(1 + i)( 1-i-/3)1 = 1(1 + i)II( l-i-/3)1= hx2 =2h

1 1 1 ... l(1+i)(1-i-/3)I = l(1+i)(1-i-/3)I =

2ii.

:. . (l ·-/3) 1

r;; [cos(i�)+isin(i�)] (1+1) l-1 3 2-v2

EXERCISE 2.05 Operations in polar form


Convert each number to polar form and hence find arg (z) and lzl.a z=l-i b z=-J3+i c z=-h+ih d z=-.!.-/ii

2 2 2 Find the argument of each complex number in radians, correct to 3 significant figures. a z=5-2i b z=-2+ 3i C z=--J3-2i d z=l-i-J5


3 Evaluate each of the following. ( 1 i) a arg 2-2

b /3[cos (2) + i sin (2)]/ c arg (cis (0) x cis (a)) d Scis(;)

.Js cis( i) 4 i&·ini·IIIN Use the theorems arg (z

1z2

••• z,,) = arg (z1 ) + arg (z2) + ... + arg (z,,)and /z

1z

2 ••• z,,/ = /z,l!z2 / ••• /z

11/ to simplify:a [ cos (2a) + i sin (2a)] [ cos (3a) + i sin (3a)] [ cos ( 6a) + i sin ( 6a)]

b 2[ cos(¾)+isin(¾) ]x.J2[ cos(¾)+isin(¾) ]x-J6[ cos(�;)+isin(�;)] c [ cos(3; )+ i sm(3;) ][ 3cos(; )-3isin(;) ][ .Js cos(-�n )-i.Js sin(-�n)]d

1

[ 4cos( il )+4i sin( il) r X 3[ cos( 3; )+i sin( 3;)]5 Use the rules /z"/ = /z/11 and /z-1

1

/ = /z/-11 to find

a /cos (0) + isin (0)/5 b 13[ cos( 2J) + i sin( 2J)] 14

6 Use the rules arg (z11

) = n x arg (z) and arg (z-1

1) = -n x arg (z) to simplify a arg {[cos (0) + isin (0)]6} b arg{[ cos(3;)+ism(3;)J} l 1 }c arg

4 [ cos ( 49n) + i sin ( 49n ) ] 7 ii•i11Hli-i First express each of the following in polar form and then simplify. Leave your answer in polar form where appropriate.

a ( .J3-i)(1+i)


8 a Express (l+i)(1+i.J3)in mod-argform. b Hence find the exact value of cos( 7! ).

-J2 +i-J2 b l+i.J3

( 2- 2i.f3")(-.f3" + i) d (4+4i)

9 Ifr[cos(a)+isin(a)]= 3-i_, find the exact value of sin (a). 2+51 I 10 If {r[ cos ( a) + isin ( a)]} 1

1 = x + yi, where x and y are real, prove that r = ( x2 + y2 )211.


tl•Id DE MOIVRE'S THEOREM

The rules in the previous section can be extended to powers of z= r[cos (0) + i sin (0)]. This work was developed by the French mathematician Abraham De Moivre (1667 -1754) and is called De Moivre's theorem.

IMPORTANT

De Moivre's theorem

If z = cos (0) + i sin (0) is a complex number, then z!' = cos (n0) + i sin (n0), V n E Z.

Proof

The proof relies on mathematical induction.

Proposition

Let P(n) be the proposition that [cos (0) + i sin (0)]" = cos (n0) + i sin (n0), V n E N, n :2: 1.

RTP

Both P(l) is true and P(k + 1) is true given that P(k) is true.

Proof

When n= l,

LHS = [cos (0) + i sin (0)]1

= cos (0) + i sin (0)

:. LHS = RHS => P(l) is true.

Assume that P(k) is true, i.e. that

RHS = [cos (1 x 0) + i sin (1 x 0)]

= cos (0) + i sin (0)

[cos (0) + i sin (0)t = [cos (k0) + isin (k0)], for some k E N, k :2: 1 .

P(k + 1): [cos (0) + i sin (0)t + 1 = cos [(k + 1)0] + i sin [(k + 1)0], for some k E N, k :2: 1.

Consider the LHS of P(k + 1):

[ cos (0) + i sin (0)t + 1 = [cos (0) + i sin (0)][cos (0) + i sin (0)t

= [cos (0) + i sin (0)] [cos (k0) + i sin (k0)], using P(k)

= cos (0) cos (k0) + cos (0) i sin (k0) + i sin (0) cos (k0) + i2sin (0) sin (k0)

= cos (0) cos (k8) - sin (8) sin (k0) + i[cos (0) sin (k0) + sin (0) cos (k0)]

= cos (0 + k0) + i sin (0 + k0)

= cos [(k + 1)0] + i sin [(k + 1)0]

= RHS of P(k+ l)

Therefore P(k + l) is true.

Therefore P(n) is true by mathematical induction.


QED

9780170250306

0 Example 16

Use De Moivre's theorem to simplify the following. c [ cos(¾)-isin(¾)J4 a [c�s (0) + i sin (0)]6 b [cos (3a.) + i sin (3a.)r8

Solution

a Apply the theorem directly using the formula [cos (0) + i sin (0)]" = cos (n0) + i sin (n0)

b Let 0 = 3a. and apply the theorem.

c First, write the expression in polar form.

Now apply the theorem and simplify.

[cos (0) + i sin (0)]6 = cos (60) + i sin (60)

[ cos (3a.) + i sin (3a.) rs

= cos (-24a.) + i sin (-24a.)

[ cos ( ¾ )-i sin ( ¾) r = [ cos ( -47t) + i sin ( -t ) r

[ cos(-47t) + i sin(-47t) r = cos(-7t )+ i sin(-7t) =-1

De Moivre's theorem is useful for evaluating powers of complex numbers. First, convert the expressions to polar form.

0 Example 17 . • . . 8 -1+i-Ji, (-1+i-Ji,) Convert --- to polar form and hence evaluate --- , giving your answer in the form

b. 2 . 2

a+ I.

Solution

First, find rand 0.

Now apply De Moivre's theorem and simplify.

Convert to Cartesian form.

9780170250306

r= (-;1 )2 +(fr =l

cos(0) = -;_1 , sin(0) = 1 ==> 0 = 237t. -1+i-Ji, (2 ) (2 ):.---=cos ; +isin ; 2

[ cos( 2;)+ i sin( 23n) ]8 = cos( 111t )+ isin( 11n)= cos( 4;)+isin( 4;)or cos(-�n) + i sin(-�n)

(-27t) .. (-27t) 1 Ji.cos -3- +zsm -3- =-2- 2 1

:.(-1+i-Ji,)8 =-.!_- Ji i.2 2 2


ws

Using De Moivre's theorem

You can now use a combination of rules to simplify expressions. 0 Example 18

. . [ cos(2p)+isin(2p)J x[ cos(3P)+isin(3P)]4

Simplify 2 [ cos(P)+isin(P)]

Solution

Use De Moivre's theorem to simplify each bracket.

U�e z1z

2 = r1r2

[cos (01 + 0

2) + i sin (01 + 02)] and5_=!!._ [cos(01 -02 )+isin(01 -02)] Z2 r2

[ cos(2p)+isin(2p)J x[ cos(3P)+isin(3P)]4

[ cos(P)+isin(P)]2

[ cos(14P )+ i sin(14P)] x [ cos(12p) + i sin(12p)] cos( 2p )+ i sin( 2p)

= cos (14P + 12p-2P) + i sin (14P + 12p -2P) = cos (24P) + i sin (24P)

EXERCISE 2.06 De Moivre's theorem


M·ini:i@iJ Use De Moivre's theorem to simplify each complex number. a [cos (0) + i sin (0)]9

b [cos (0) + i sin (0)r2

c [cos (0) -i sin (0)]4 d [cos (20) + i sin (20)] 3

2 Evaluate z4 in Cartesian form if a z = 2[cos(¾)+isin(¾)] c z= }i[ cos(i)+isin(i)]

3 E 1 [ (-37t) .. (-37t)]-6va uate cos -5- +ism -5-

b z=3[cos(%)+isin(%)] d z = s[ cos(i;)+isin(i;)]

4 i%•1:,j.jlr4 By first converting to polar form, evaluate each of the following, giving your answerin Cartesian form. 6 a (1 + i)3 b ( ._13 + i r C ( h - ihf d ( -� i r5 Evaluate the following.

1 a (2+2i)6

6 Show that(Jz + Jzif (½+ �ir =l

1 b---(1-i"'3)8


7 •%-ini-lif:j Simplify each complex number. a [cos (0) + i sin (0)] 5 x [cos (0) + i sin (0)r8

[ cos ( 4<j>) + i sin ( 4<j>) J5

C

[ cos(7<1> )+i sin(7<!>) J3

b [cos (a)+ i sin (a)] 3 x [cos (P) + i sin (P)l4

[ cos(P )- i sin(p) ]2 x [ cos(P) + i sin(P) T3 d--------------

[ cos( 3P )+ i sin( 3P) ]4

Reasoning and communication 1 8 If z == cos (0) + i sin (0), prove that z2

+ 2 is always real.z

1 9 If z == cos (0) + i sin (0), prove that z3 - 3 is purely imaginary.

z

10 If z == cos (0) + i sin (0), what is the value of z" + _!__? z"

tl•t4 APPLICATIONS OF

DE MOIVRE'S THEOREM

De Moivre's theorem is useful for deriving or proving a number of trigonometric identities. These will be examined in this section.

0 Example 19

Use the binomial theorem to expand [cos (0) + i sin (0)] 3. Hence derive a formula for cos (30) in terms of cos (0).

Solution

Recall the expansion for (a+ b)3. [cos (0) + i sin (0)] 3

== cos3(0) + 3 cos\0) [i sin (0)] + 3 cos (0) [i2sin\0)] + i3sin3(0)

== cos3(0) + 3i cos\0) sin (0) - 3 cos (0) sin2(0) - i sin3(0)

Use De Moivre's theorem to find an expression [cos (0) + i sin (0)] 3 == cos (30) + i sin (30) with cos (30).

Equate the real parts. Use a substitution to eliminate sin\0).

9780170250306

:. cos (30) + i sin (30) == cos3(0) + 3i cos2(0) sin (0)

- 3 cos (0) sin\0) - i sin3(0):. cos (30) == cos3(0) - 3 cos (0) sin\0) cos (30) == cos3(0) - 3 cos (0)(1 - cos\0)]

== cos3 (0) - 3 cos (0) + 3 cos3(0) == 4 cos3(0) - 3 cos (0)

:. cos (30) == 4 cos3(0) - 3 cos (0)


0 Example 20

By expanding [cos (a)+ i sin (a))2 in two ways, derive expressions for cos (2a) and sin (2a) and hence find an expression for tan (2a) in terms of tan ( a).

Solution

Use the binomial theorem.

Use De Moivre's theorem.

Equate the real and imaginary parts.

sin(A) Use tan( A)= -

cos( A)

Convert to tan (a) by dividing

h HS b cos2 (a) t eR y---.cos2 (a)

[cos (a)+ i sin (a)J2 = cos2(a) + 2i cos (a) sin (a)+ i2sin2(a) = cos2(a) - sin2(a) + 2i cos (a) sin (a)

[cos (a)+ i sin (a))2 = cos (2a) + i sin (2a)

cos (2a) = cos2(a) - sin2(a) sin (2a) = 2 sin (a) cos (a)

sin(2a) tan(2a)=-��cos(2a)

_ 2sin( a ) cos( a) -

cos2(a)-sin2 (a) 2sin(a)cos(a)

tan(2a)= cos2 (a)

cos2 (a) sin2(a) -----

cos2

(a) cos2 (a)

= 2 tan(a)l-tan2(a)

Expansions using De Moivre's theorem and the binomial theorem can also be used to integrate powers of trigonometric functions.

0 Example 21

7t

Use the fact that cos (30) = 4 cos3(0) - 3 cos (0) to find f 02 4cos3 (0)d0.

Solution

Rearrange cos (30) = 4 cos3(0) - 3 cos (0) to make 4 cos3(0) the subject.

cos (30) = 4 cos3(0) - 3 cos (0) :. 4 cos3(0) = cos (30) + 3 cos (0)

Perform the integration.


7t 7t

f 0

2 4cos3 (0)d0 =ft cos(30)+ 3cos(0)d07t

[sin(30) . (0)]2

= --+3sm 3

0

sin[ 3(�)] . (n) [

sin[ 3(0)] . ( )l = 3 + 3 sm 2 - 3 + 3 sm O

1 =--+3(1)-(o) 3

=2�3

9780170250306

INVESTIGATION Euler and e;e

Leonhard Euler was born in Switzerland in 1707. His father wanted him to study Theology, but Euler persuaded his father to allow him to study Mathematics. Euler was the first mathematician to use the notationf(x) for a function and in 1748 he defined the formula /8 = cos (0) + i sin (0). This definition was very useful in proving the various results to do with complex numbers, including De Moivre's theorem. By expressing the complex number z= r[cos (0) + i sin (0)) as z= re;e, prove the following results.

z1z2 . . . z11 = r1r2 ... r11

[cos (01 + 02 + ... + 0

11) + i sin (0

1 + 02

+ ... + 011))

2 z1' = r11[cos (n0) + i sin (n0)], 'v' n E Z

3 ..:i= !L[cos(01 -02

)+isin(01 -02 )]

Z2 r2

4 z-1 = _!_ = .!. [cos(-e)+isin(-0)] z r

EXERCISE 2.07 Applications of De Moivre's theorem


'4H,,i,ji@ a Expand [cos (0) + i sin (0)) 3 using the binomial theorem. b Use the fact that [cos (0) + i sin (0)) 3 = cos (30) + i sin (30) to derive an expression for

i cos (30) in terms of cos (0) ii sin (30) in terms of sin (0)

2 a Expand [cos (0) + i sin (0))2 in two ways. b Hence show that

i cos (20) = 2 cos2 (0) - 1 ii cos (20) = 1 - 2 sin2 (0)

3 Use expansions of [cos (0) + i sin (0)]4 to prove that a cos (40) = 8 cos4 (0) - 8 cos2 (0) + 1 b sin (40) = 4 sin (0) cos (0)[cos2 (0) - sin2 (0))

Trigonomelric identities using De Moivre's theorem


4 -#•1:,1•iiM a Complete the statement tan(30) = _(_(39)).cos --

b Using tlie'iesuJts you derived from [cos (0) + i sin (0)]3 = cos (30) + i sin (30) in question 1, . h

'.-c'_ .. -'h·· .• · · ·_ ..

(· ·0)

' 3sin(0)-4sin3(0)s ow t a.t tan 3 = 3 . ,. ·. ,.- · . . 4cos (0)-3cos(0).. _· . : , 3 tan ( 0 )-tan 3 ( 0) c Hence prove that tan(3e) =.

2 -. · ·. ·. l-3tan (0)

5 For n :C:: 1, prove that [ cos ·(0) + i sin (0)] + [ cos (20) + i sin (20)] + [ cos (30) + i sin (30)] + ... + [ cos (n0) + i sin (n0)] _ [ cos(0)'+ isin(0) ][ cos(n0)+ i sin(n0)-1)

- cos(0) + i sin(0)-1. _ . . . . cis( 60°) 6 If 60° -45° = 15°, express c1s ( 60°) and c1s ( 45°) 111 Cartesian form and hence simplify . ( ). C!S 45°

111 two ways. Use your results to find the exact value of cos (15°). 7 M·lnHIU a Use [cos (0) + i sin (0)]3 = cos (30) + i sin (30) to show that

sin (3a) = 3 sin (a) -4 sin3(a). n

b Hence find the exact value of J: 4sin3 (0)d0. 2,r

8 Using suitable expansions, find the value of J,:3 cos4 (x )dx.3


9 Prove by induction that [cos (0) + i sin (0)]'1 = cos (n0) + i sin (n0), n :C:: l, n E Z.

10 If z = cos (0) + i sin (0), simplify 1 2 1 a z-- b z --z z

2


1C Z

11

--

z"

9780170250306

P(x,y) y ------------r

I

I

I

I

I -----70-+------�x----:

x

Im (z)

P(z)

0 Re(z)

9780170250306

- • •

arg (z) = 8

If a complex number is represented by a

point P in the complex plane, then the

argument of z, denoted arg (z), is the angle

8 that OP makes with the positive real axis

Ox, with the angle measured anticlockwise

from Ox. The principal value of the

argument is the one in the interval (-n, n].

The argument of O is not defined.


0 x = rcos (9) X

Let arg (z) = 0 and lzl = r.If z is a non-zero complex number, then z= r[cos (0) + i sin (0)] is the polarform ofz.

• z= r[cos (0) + i sin (0)] is also written asz= r cis (0)

• If z1 = ri[cos (01) + i sin (01)] and

z2 = r 2 [ cos ( 02) + i sin ( 02)] are two complex numbers, then

• their product isz

1z

2 = r1r2 [cos (01 + 02) + i sin (0

1 + 02)]

• arg (z1z

2) = arg (z1) + arg (z

2) andmod (z

1z

2) = mod (z

1) x mod (z

2)

II If z1 = ri

[cos (01) + i sin (01)] and

z2 = r2 [cos (02) + i sin (02)] are two

complex nm11bers, then .:!.. = l[ cos(01

- 02)+isin(01 -82 )].

Z2 r2

If z= r[cos (0) + i sin (0)] is a complex number, then z -I = 2. = � [ cos ( -0) + i sin ( -0)].

z r OR

If z1 = r1[ cos (01) + i sin (01)] and

z2 = r2 [cos (02) + i sin (0

2)] are two complex numbers, then arg( :: J = arg( z1 )- arg( z2 ) andmod(.:!..J = mod{zi ).

z2

mod(z2 )


complex numbers, then z

1z

2 ••• z,, = r1r2 ... r11 [cos (01

+ 02 + ... + 811

)+ i sin (01 + 02 + ... + 811)].

OR

arg (z1z

2 • • • z

11)

= arg (z1) + arg (z2) + ... + arg (z,,) andlz1z2 · · · z"I = lzillz2I · · · lz,,I.

• arg (z") = n x arg (z) and lz"I = lzl"andarg (z-11) = -n x arg (z) and lz-"I = lzl-".

• De Moivre's theorem

If z = cos (0) + i sin (0) is a complex number,then z" = cos (n0) + i sin (n0), V n E Z.

• De Moivre's theorem is useful for deriving orproving a number of trigonometricidentities.• Expansions using De Moivre's theorem andthe binomial theorem can also be used tointegrate powers of trigonometric functions.

9780170250306

CHAPTER REVIEW

COMPLEX NUMBERS AND

DE MOIVRE'S THEOREM

Multiple choice

;;.;,,;.;a The roots of x2 - 2x + 10 = 0 are

A -1 ± 3i B l±i-,J3D -1 ± i-,J3 E none of the above

2 •%l•1,:1•1JI The value of i74 is: A 1 B -1 C

3 •%•1,:j,jiij The value of3- 2i + i ( 5 + 4i) is: A 1 -3i B -1 + 3i C -1 - Si

C 1 ±3i

D -i E 74

D -1-3i E 7- 3i

4 ;;;.;,,;.;a Consider the vectors u and v representing the complex numbers u and v respectively, as shown in the diagram.

Im (z)

u

0 Re(z)

V

Which of the following statements is true? A u = V B u = -v C - u = -v o u=v

5 M•1,:j.j@11 The principal value of the argument of2cis(3�n) is:

A Sn6

B -n6

C 77t

6 D n

6

E u=v

E sn6

9780170250306 CHAPTER 2: Complex numbers and De Moivre's theorem 1101

Short answer

6 i#•1111•14 Simplify (3 - 2i)(l + 4i) -(2 + 5i)2•

7 iaj,j,ij.jp Realise the denominator of 5 + � .2-t 8 ii·'H""ill Express ( 2-i.f3( in the form a+ bi. Hence find:

a Re[(2-i.f3(] b 1m[(2-i.f3(]9 ij!j\\j,,?J Write down the complex number z in mod-arg form given that arg ( z) = �; and

1011

mod( z) = 2.Js.

12 ;;.,,.,.,lit [f z = 2[ cos( 3;)+isin(3;)J, w = 3[ cos( 5;)+isin( 5;)] and u =5[ cos(1�)+isin(�)],find a arg (zwu) b Jzwul

Application

13 Evaluate ( .}i-� J, giving your answer in Cartesian form.14 Write the expansions of

a cos (A± B) b sin (A± B).Hence prove thatc [cos (0) + i sin (0)] [cos(;\)+ i sin(;\)]= cos (0 + ;\) + i sin (0 + ;\) d [cos (0) -i sin (0)] [cos(;\) -i sin(;\)]= cos (-0 - ;\) + i sin (-0 - ;\)

15 Use De Moivre's theorem and the binomial theorem to expand [ cos (0) + i sin (0) ]3 in twoways. Hence, or otherwise, find the value of J 2 cos3 (Sa) da. Give your answer correct to 3-1 significant figures.

16 Evaluate { 2[ cos(3; )-i sin(3;) ]f 5, giving your answer in Cartesian form.17 Prove the identity Jz1z21 = Jz1 1Jz2J for any two complex numbers Zp z

2• Hence prove by

mathematical induction that Jz1z2z3

• • . z11 I = lz

1 I Jz

2J lz31 ... lz

11I for all positive integers n.

18 Simplify each of the following.1

cos(Sx )+ i sin(Sx)3[ cos(;)+isin(;)]b 4[ cos(� )+isin(�)]

19 Express -h-ih in mod-arg form. Hence find the value of:

9[ cos(�)+isin(�)]C 3[ cos( 4�) + i sin( 4�)]

a arg(-h- ih( b mod(-h-ih(


20 Find the value of each of the following in mod-arg form. a ( ·F3)( F3 ·) b [(-1-iF3)(2+2i)]-1

4-41 3 - 3+1

21 Simplify each of the following.[ cis(�)J[ cis(¥)J5

a 6 [ cis(2!)]

[ cos ( 4a) + i sin ( 4a) ]7

b-- ----'�-------"---- -[ cos(9a)+i sin(9a)r4 [ cos(9a)-isin(9a)]3

22 By expanding [cos (0) + i sin (0)]2 in two different ways, derive expressions for cos (20) and sin (20). Hence show that: 2tan(a) a tan(2a) = -----"-----'--- b tan(15° ) = 2-F3 l-tan2(a)

Qz

Practice quiz

9780170250306 CHAPTER 2: Complex numbers and De Moivre's theorem 1103

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