De Moivre S Theorem and Its Applications

UNIT I

DE MOIVRE’S THEOREM AND ITS APPLICATIONS

DE MOIVRE'S THEOREM AND ITS APPLICATIONS

Learning objectives:

1. To understand the origin of DeMoivre’s theorem

2. To study the definition and proof of DeMoivre’s theorem

3. Employ DeMoivre’s theorem in a number of applications such as:

 Raising a complex number to a power.

 Finding roots of complex numbers.

1  Positioning the roots of a complex number on an Argand diagram.

 Finding the nth roots of a complex number.

 Representing a complex number in exponential form.

 Expressing Cosn, Sinn and tann in terms of Cos, Sin and tan.

Introduction

In this module we introduce De Moivre’s theorem and some of its applications. De Moivre is a famous French Mathematician from the

1700's. He was a friend of Isaac Newton and was famous for his work with complex numbers, trigonometry and the theory of probability.

De Moivre's theorem links complex numbers and trigonometry.

Let z1 and z2 be two complex numbers. Then they can be represented as

z1= r 1( Cos q 1 + i Sin q 1 )

z2= r 2( Cos q 2 + i Sin q 2 )

Where z1= r 1 , z 2 = r 2 , arg( z 1) =q 1 . arg(z 2 ) =q 2

2 Now z1 z 2= r 1 r 2( Cos q 1 + i Sin q 1)( Cos q 2 + iSin q 2 )

r r轾 Cosq Cos q + i2 Sin q Sin q + iCos q Sin q + iSin q Cos q = 1 2臌 1 2 1 2 1 2 1 2

= r1 r 2臌轾 Cosq 1 Cos q 2 - Sin q 1 q 2 + i( Cos q 1 Sin q 2 + iSin q 1 Cos q 2 )

z1 z 2= r 1 r 2臌轾 Cos( q 1 + q 2) + iSin ( q 1 + q 2 )

\z1 z 2 = r 1 r 2 , arg( z 1 z 2) =( q 1 + q 2 )

Inference

\z1 z 2 = z 1 z 2 , arg( z 1 z 2) = arg( z 1) + arg( z 2 ) ie.,

(1) The modulus of the product of two complex numbers is the product of

their moduli.

(2) The argument of the product of two complex numbers is the sum

of their arguments.

Using these facts we can find the square of a complex number (in polar form).

Let z= r( Cos q+ iSin q) where z = r and arg(z) =q

Then z2= r 2 ( Cos q+ iSin q)2

2 2 2 2 = r臌轾 Cosq+ i Sin q+ 2iSin q Cos q

3 2 2 2 = r臌轾 Cosq- Sin q+ i2Sin q Cos q

= r2 [ Cos 2q+ iSin 2 q]

It is clear that z2= r 2 and arg( z 2 ) = 2 q

This concept can be used to find any positive integer power of z.

De Moivre's Theorem

Let n be any rational number, positive or negative, then

(Cos +i Sin )n = Cos n + i Sin n 

Case(i) when n is a positive integer. We prove the theorem by mathematical induction. i.e. we wish to show that

(Cos  + iSin )n = Cos n  + i Sin n

When n = 1 we have

(Cos  + i Sin )1 = Cos 1 + i Sin 1 = Cos  + i Sin

Which is true, so the theorem is true for n=1

Now assume that the theorem is true for n=k, where k is a non-negative integer. Then we have.

(Cos  + i Sin )k = Cos k  + iSin k

4 Multiplying both sides by (Cos  + i Sin ) we get

(Cos  + i Sin )k+1 = (Cos k + i Sink) (Cos  + i Sin)

= (Cos (k) Cos  - Sin (k) Sin ) + i (Sin(k) Cos + Cos (k) Sin

= Cos (k + ) + i Sin (k + )

= Cos (k+1)  + i Sin (k+1) 

So, if the theorem is true for n=k, it is true for n=k+1. Hence by the principle of mathematical induction the theorem is true for all n.

Case (ii) When n is a negative integer

Let n = -m and m > 0

Now (Cosq+ iSin q)n =( Cos q+ iSin q)- m

1 = m (Cosq+ iSin q)

1 = (Cos mq+ iSin m q)

Cos mq- iSin m q = (Cos mq+ iSin m q)( Cos m q- iSin m q)

Cos mq- iSin m q = Cos2 mq- i 2 Sin 2 m q

5 Cos(mq ) - iSin m q = Cos2 mq+ Sin 2 m q

= Cos( mq) - iSin (m q )

= Cos(- m q) + iSin( - m q)

= Cos nq+ iSin n q

Case (iii) When ' n ' is a fraction

p Let n = where p is any positive integer and q any non zero integer. q

From case (1) we have (Cosq+ iSin q)p =( Cos p q+ iSin p q)

p p = Cos qq+ iSin q q q q

q 骣 p p = 琪Cosq+ iSin q 桫 q q

Now taking qth root of both sides, (Cos  +Sin ) p/q has one of its value

p p = Cosq + iSin q q q

p n Now writing =n we get( Cos q+ iSin q) = Cos n q+ iSin n q q

Hence the theorem.

6 Application of De Moivre's Theorem

De Moivre’s theorem is an important result in the field of complex numbers and it has practical applications.

1. To find the positive powers of a complex number.

Example: Find z10 if z = 1-i

Let 1-i = r (Cos + i Sin )

Equating real and imaginary parts we get

r Cos  = 1 r Sin  = -1

r2(Cos2 + Sin2) = 1+1

r2 = 2

 r = 2

rSin q = -1 tan  = -1 r Cos q

-p   = 4

7 -p arg (z) = 4

� 骣p- 骣 p \1 - i = 2犏 Cos琪 + iSin 琪臌桫4 桫 4

Applying De Moivre's theorem we get

10 10 � 骣p- 骣 p (1- i) =( 2) 犏 Cos 10琪 + iSin10 琪臌桫4 桫 4

5 � 骣p5 - 骣 p 5 = 2犏 Cos琪+ iSin 琪臌桫2 桫 2

� 骣p5 - 骣 p 5 = 32犏 Cos琪+ 2 p + iSin 琪 + 2 p 臌桫2 桫 2

� 骣p- 骣 p = 32犏 Cos琪+ iSin 琪臌桫2 桫 2

閜骣-p = 32犏 Cos- iSin 琪臌 2桫 2

= 32( 0+ i( - 1))

(l-i)10 = -32i

2. To find the nth roots of a complex number

Let z= r( Cos q+ iSin q)

8 = then z1/n= r 1/n [ Cos q+ iSin q]1/n

1/ n 靟轾+2k p 轾 q + 2k p = r睚 Cos犏+ iSin 犏 where k is an integer. 铪臌n 臌 n

Since Sine and Cosine functions are Cyclic and repeat with every 2.

To get the n different roots of z, one only needs to consider values of k from 0 to n-1.

Example: Find the three cube roots of unity.

1 = 1 + i 0

1 = 1 (Cos 0 + i Sin 0)

3 3 � 骣0p 2 + k 骣 0 p 2 k 1= 1犏 Cos琪 + iSin 琪臌桫3 桫 3

k = 0, 1, 2

= Cos 0 +i Sin 0 when k=0,

2p 2 p Cos+ iSin when k = 1, 3 3

4p 4 p and Cos+ iSin when k = 2 3 3

-1 3 - 1 3 = 1, +i , - i 2 2 2 2

Example

9 Find all complex roots of 27i

27 i = 0 + 27i

27i= 02 + 27 2 = 27

p rSin = 27 arg (27i) = 2 27 Sin  = 27 骣 p p  Sin  =1 27i = 27琪Cos+ iSin 桫 2 2 Hence  = /2

Let z = r (Cos  + i Sin ) So that z3 = 27i

By De Moivre’s theorem

r3 (Cos 3 + i Sin 3 ) = 27i

= 27 (Cos /2 + i Sin/2)

 r3 = 27  r = 3

Cos 3 = Cos /2, Sin 3 = Sin /2

3 = /2

Hence 3 = /2 + 2k

Case(i) when k =0

3 = /2

10  = /6

z1 = 3(Cos /6 + i Sin /6)

3 3 3 = 3 (3/2+i ½ ) = + i 2 2

Care (ii) When k = 1

p 3q= + 2 p x1 2 5p 3q= 2 骣 5p p \z2 = 3琪 Cos + iSin 桫 6 6

骣- 3 1 = 3琪 +i 桫2 2

-3 3 3 z= + i 2 2 2

Case (iii) When k = 2

p 3q= + 2 p x 2 2

p = +4 p 2

9p 3q= 2

骣 3p 3 p \z3 = 3琪 Cos + iSin 桫 2 2

11 = 3 (0+i(-1)

z3 = -3i

 The complex cube roots of 27i are

3 3 3 z= + i 1 2 2

-3 3 3 z= + i 2 2 2

z3 = -3i

Example 2:

Find all complex fourth roots of -16

-16 = -16 + i0

= r( Cosq+ iSin q)

r Cos  = -16 r Sin  = 0

r2( Cos 2q+ Sin 2 q) = 256

r2 = 256  r = 16

Now 16 Cos  = -16

-16 Cos  = 16

12 Cos  = -1

  = Cos-1(-1) = 

 -16 = 16 (Cos  + i Sin )

Fourth roots of 16 have modulus

4 16= 2 and the possible arguments are

p p +2 p 3 p p + 4 p 5 p p + 6 p 7 p ,= , = , = 4 4 4 4 4 4 4

Hence fourth roots of -16 are:

閜骣 p z1 = 2犏 Cos琪 + iSin 臌桫4 4

= 2+ 2 i

轾 3p 3 p z2 = 2 Cos + iSin 臌犏 4 4

= - 2+ 2 i

轾 5p p z3 = 2 Cos + iSin 臌犏 4 4

= - 2- 2 i

轾 7p 7 p z4 = 2 Cos + iSin 臌犏 4 4

13 = 2- 2 i

3. To prove trigonometric identities

Certain trigonometric identities can be derived using DeMoivre's theorem. We can express Cos n, Sin n  and Tan n in terms of Cos , Sin  and Tan .

Example

1. Prove that Cos 5 = Cos5 - 10 Cos3 Sin2 + 5 Cos  Sin4

2. Prove that Sin 5 = 16 Sin5 - 20Sin3 + 5Sin

tanq- 10 tan3 q+ tan 5 q 3. Prove that tan 5 = 1- 10 tan2 q+ 5tan 4 q

Solution:

1) Let C = Cos  and S = Sin 

Cos 5 = Real part of (Cos  + iSin )5

= Re (C + iS)5

5 4 3 2 2 3 4 5 = Re (C + 5C1C (iS) + 5C2 C (iS) + 5C3 C (iS) + 5C4 (iS) + (iS)

= Re (C5 + 5C4 iS + 10 C3i2S2 + 10 C2i3S3 + 5Ci4S4 + i5S5)

= Re (C5 + 5C4Si+10C3(-1) S2 + 10C2(-i) S3 + 5C(1) S4 + iS5)

Cos 5 = Re(C+5C4Si-10C3S2-10C2S3i + 5CS4 + iS5)

14 = C5 – 10C3S2 + 5CS4

= Cos5 - 10 Cos3  Sin2  + 5Cos Sin 4

It is to be noted that, the right hand side can be expressed

entirely in terms of Cos  by substituting Sin2 = 1- Cos2 and Sin4 =

(1 - Cos2)2

 Cos 5 = Cos5 - 10 Cos3 (1-Cos2) + 5 Cos (1-Cos2)2

= Cos5 - 10Cos3 + 10 Cos5 + 5 Cos(1-2 Cos2 + Cos4)

= Cos5 - 10 Cos3 + 10Cos5 + 5 Cos - 10Cos3 + 5Cos5 

Cos 5 = 16 Cos5 - 20 Cos3 + 5 Cos

2. Sin 5 = Imaginary part of (Cos 5 + i Sin 5)

= Im (Cos  + i Sin)5

= Im (C + is)5

= Im (C5 + 5C4 Si – 10C3S2 – 10C2 S3i + 5CS4 + is5)

= 5C4 S-10C2S3 + S5

 Sin 5 = 5 Cos4  Sin  - 10 Cos2Sin3 + Sin5 

If required, the right hand side can be expressed entirely in terms of

Sin  by substituting Cos2 = 1-Sin2, and Cos4 = (1-Sin2)2

15 Sin 5 = 5 (1-Sin2)2 Sin-10 (1-Sin2) Sin3 + Sin5 

= 5(1-2Sin2 + Sin4) Sin  - 10(1-Sin2) Sin3 + Sin5

= 5 Sin-10Sin3+5Sin5-10Sin3+10 Sin5+Sin5

= 16 Sin5 - 20Sin3 + 5Sin

Sin 5q 5Cos4 q Sin q- 10Cos 2 q Sin 3 q+ Sin 5 Q tan 5 = = Cos5q Cos5q- 10Cos 3 q Sin 2 q+ 5Cos q Sin 4 q

Dividing every term on Numerator and denominator by Cos5 we get

tanq- 10 tan3 q+ Tan 5 q Tan 5q= 1- 10 tan2 q+ 5tan 4 q

3. Exponential form of a complex number.

If  is measured in radians both Cos and Sin  can be expressed as an infinite series in powers of  as follows:

q2 q 4 q 6( - 1) n- 1 q 2n - 2 Cosq= 1 - + - + ... + + ... .and 2! 4! 6! (2n- 2)!

q3 q 5 q 7( - 1) n- 1 q 2n - 1 Sinq=q- + - + .... + + .... 3! 5! 7! (2n- 1)!

The exponential series is given by

x x2 x 3 x n- 1 ex = 1 + + + + . ... + + .. ... 1! 2! 3! (n- 1)!

16 If we replace x by i, we get

(iq)2( i q) 3( i q) n- 1 eiq = 1 + i q+ + + ... + + ... 2! 3! (n- 1)!

q2i q 3 q 4 q 5 = 1+ i q- - + + i + ...... 2! 3! 4! 5!

骣q2 q 4 骣 q 3 q 5 = =琪1 - + - ... + i 琪 q - + .... 桫2! 4! 桫 3! 5!

eiq = Cos q+ iSin q , it is known as Euler’s formula

It is to be noted that if z = Cos+iSin then zn = (Cos + i Sin)n

= Cosn + i Sin n 

= eni

and if z = r (Cos  + i Sin) then

z = rei and zn = rneni

The form rei is known as the exponential form of a complex number and is linked to polar form very clearly.

Another result that can be derived from the exponential form of complex number is the following.

eiq = Cos q+ iSin q (1)

17 e-i q = Cos( -q) + iSin ( -q)

e-i q = Cos q- iSin q (2)

(1) + (2) 

 ei + e-i = 2Cos 

eiq+ 2 - 1 q \Cos q= 2

(1) - (2) 

 ei - e-i = 2 i Sin 

Additional Problems:

3 (1) 骣 p Simplify 琪Cosp / 6 + iSin 桫 6

3 骣 p p3 p 3 p 琪Cos+ iSin = Cos + iSin 桫 6 6 6 6

p p = Cos+ iSin 2 2

= 0 + i

(2) Find the powers of (3+ i)3

3+ i = r( Cos q+ iSin q)

r Cosq= 3 rSin q= 1

18 2 r2( Cos 2q+ Sin 2 q) =( 3) + 1 2

= 3 + 1

r2 = 4

\r = 4 = 2

rSin q 1 = r Cosq 3

1-1 骣 1 tanq= \q= tan 琪 3桫 3

p q= 6

3 3 閜骣 p \( 3 + i) =犏 2琪 Cos + iSin 臌桫 6 6

3 轾 3p 3 p = 2 Cos+ iSin 臌犏 6 6

骣 p p = 8琪 Cos+ iSin 桫 2 2

= 8(0 + i)

(3) Show that Cos 3 = 4 Cos3 - 3 Cos 

Cos 3 + i Sin 3 = (Cos  + i Sin )3

= (C + iS)3

= C3 + 3C2i S + 3C i2S2 + i3S3

= C3 + 3C2 iS – 3CS2 – iS3

19 Cos 3 + i Sin 3 = (C3 – 3CS2) + i (3C2 S – S3)

Equating real and imaginary parts on both sides we get

Cos 3 = C3 – 3CS2

= Cos3 - 3 Cos  Sin2

= Cos3 - 3 Cos (1-Cos2)

= Cos3 - 3 Cos + 3 Cos3

= 4 Cos3 - 3 Cos

4. Express tan 4 in terms of tan 

Sin 4q tan 4 = Cos 4q

Sin 4 and Cos4 can be expressed in terms of Sin  and Cos . Using

DeMoivre’s theorem we have

Cos4 + iSin4 = (Cos + i Sin)4 1

= (C + iS)4 1 1

= C4 + 4C3iS+6C2(iS)2 + 4C(iS)3 + (iS)4 1 2 1

= C4 + 4C3iS + -6C2S2 – 4 i CS3 + S4 1 3 3 1

Cos4 + i Sin4 = (C4 – 6C2S2 + S4) + i (4C3S – 4CS3) 1 4 6 4 1

Equating real and imaginary parts we get

Cos 4 = C4 – 6C2S2 + S4

= Cos4 - 6 Cos2 Sin2 + Sin4

20 Sin 4 = 4Cos3 Sin  - 4Cos Sin3

Sin 4q tan 4 = Cos 4q

4Cos3q Sin q- 4Cos q Sin 3 q tan 4 = Cos4q- 6Cos 2 q Sin 2 q+ Sin 4 q

Dividing every term by Cos4, we get

Sinq Sin3 q 4- 4 Cosq Cos3 q Tan 4 = Sin2q Sin 4 q 1- 6 + Cos2q Cos 4 q

4 tanq- 4 tan3 q = 1- 6 tan2 q+ tan 4 q

1. The modulus of Z = 3-4i is ______

a) 3 b) -3 c) 5 d) 7

2. The argument of the complex number -1 + i 3 is ______

2p 2p 2p 2p a) b) c) d) 3 3 3 3

3. The value of eip is ______

a) 2 b) -2 c) -3 d) -1

6 4. If ( 3+ i) = x + iy then the values of x and y are ______,

______respectively.

a) x = -64, y = 0 b) x = -60, y = 1

c) x = -64, y =3 d) x = 64, y = -5

21 z= r cos a + isin a and Z= s Cos b+ iSin b Z1 5. If 1 ( ) 2 ( ) then the Z2 is ______

s s a) 轾Cos(a -b) + iSin( a -b) c) 轾Cos(a +b) + iSin( a +b) r 臌 r 臌

r s b) 轾Cos(a -b) + iSin( a -b) d) 轾Cos(a -b) + iSin( a -b) s 臌 r 臌

1+ C + iS 6. Given C2+ S 2 = 1, then the value of is ______1+ C - iS

a) Cosq+ iSin q b) Cosq+ iSin q

c) Cosq+ iSin q d) Cosq+ iSin q

骣2+ i 7. The value of (1+ i)琪 is ______桫3+ i

a) 2 b) 1 c) 4 d) -1

1 3 8. If Z= + i then the polar form of Z2 is ______2 2

2p 2 p 2p 2 p a) Cos+ isin b) Cos+ isin 3 3 3 3

2p 2 p 2p 2 p c) Cos+ isin d) Cos+ isin 3 3 3 3

22 Ans:

2p 2. 3

4. x = -64, y = 0

r 5. 轾Cos(a -b) + iSin ( a -b) s 臌

6. Cosq+ iSin q

2p 2 p 8. Cos+ isin 3 3

1. Expand (1+ i)10 in the form of a+ib.

2. Convert the complex number 1+ 3 i to polar form.

cos 2q+ i sin 2 q 3. Simplify the expression cos3q+ i sin 3 q

n 轾1+ Cos q+ iSin q 4. Prove that =Cosn q + iSinn q 臌犏1+ Cos q - iSin q

1. 32i

p p 2. 2(cos+ i sin ) 3 3

23 3. cosq- i sin q

4. 4cos3 q- 3cos q

Assignment

1. Find the cube roots of 1+i

2. Express Sin 3 in terms of Sin 

3. Prove that Cos 4 = 8 Cos4 - 8 Cos2 + 1

4. Express Sin 5 in terms of Sin

5. Express tan 3 in terms of tan 

6. Show that Cos 6 = Cos6 - 3 Cos4 + 3 Cos2 - 1

Glossary

1. The modulus - argument form of a complex number z is expressed

as z = r (Cos + iSin), where r is the modulus and  is the argument.

2. The modulus of the product of two complex numbers is the

product of their moulii.

3. The argument of the product of two complex numbers is the sum of

their arguments.

4. De Moivre’s theorem states that

n (Cosq+ iSin q) = Cosn q+ iSinn q where n is rational number

positive or negative.

24 5. The cube root of unity are 1,-1 + i3 , - 1 - i 3 2 2 2 2

6. eiq = Cos q + iSin q is known as Euler’s formula

Reference Books

Engineering Mathematics by K.A. Stroud, 6th edn.

Text book of De Moivre’s Theorem by A. K. Sharma DPH Mathematics Series.

Summary

De Moivre’s theorem is one of the important theorems in mathematics. This theorem connects complex numbers and trigonometry. One can use the generalized formula

n (Cosq+ iSin q) = Cosn q+ iSinn q to find explicit expressions for the nth roots of unity. It can be used to obtain relationships between trigonometric functions of multiple angles and powers of trigonometric functions.