EdExcel FP3 Complex numbers - rousemaths

Edexcel Further Pure 2 Complex Numbers Section 2: de Moivre’s theorem Notes and Examples These notes contain subsections on:  De Moivre’s Theorem.  Using de Moivre’s Theorem to find powers of complex numbers  Using de Moivre’s theorem to find trigonometric identities  Roots of unity  General nth roots

De Moivre’s theorem De Moivre’s Theorem states that

 cos  i sin  

n

 cos n  i sin n for any integer n.

For positive integers this is very obvious from the rules you for multiplying complex numbers in polar form (“add arguments, multiply moduli”). That it is also true for negative integers is a consequence of the facts that

 cos   i sin  

1

1 cos   i sin  cos   i sin    cos   i sin   cos   i sin   

cos   i sin  cos 2   sin 2   cos   i sin   cos( )  i sin( ) 

and  cos   isin  

n



 cos   isin  

.

1 n

De Moivre’s theorem can be proved using proof by induction, but it can be proved more easily using the exponential form of complex numbers, as shown in the textbook. You may be asked to give this proof in an examination. For practice in applying de Moivre’s theorem, try the de Moivre hexagonal puzzle.

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Edexcel FP2 Complex nos Sec. 2 Notes & Examples Using de Moivre’s theorem to find powers of complex numbers The simplest application of De Moivre’s theorem is to evaluate powers of complex numbers very quickly. Notice that, even though De Moivre’s theorem itself refers to a complex number with modulus 1, it can be used for any complex number because, when expressed in polar form, a complex number is written as a real number multiplied by a complex number with modulus 1. In polar form a complex number is written as follows:

z  x  yi  r  cos   isin   where r  z and   arg( z) . so that

z n   x  yi    r  cos   i sin    n

n

 r n (cos   i sin  )n  r n (cos n  i sin n ) This is illustrated in the examples below. Example 1

   Find the value of  cos  i sin  . 3 3  9

Solution

 , 3 9        cos  i sin    cos  9    i sin  9   3 3 3 3     cos 3  i sin 3  1

Using de Moivre’s theorem with n = 9, θ =

sin 3  0 and cos3  1

De Moivre’s theorem can only be used for complex numbers written in modulus-argument form. Complex numbers given in rectangular form must first be converted into modulus-argument form.

Example 2 Find the value of (1 + i)16. Solution Initially convert 1 + i into modulus, argument form. 1  i  2(cos 4  isin 4 )

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Edexcel FP2 Complex nos Sec. 2 Notes & Examples  (1  i)16 





2(cos 4  i sin 4 )

16

Using de Moivre with n = 16 and   4

 cos 4  i sin 4   256  cos 16  4   i sin 16  4    [ 2]

 16



16

sin 4  0 and cos 4  1

 256(cos 4  i sin 4 )  256

Always check that the complex number you are working with is in the appropriate form. In Example 3 you need to rewrite the complex number.

Example 3 6 Find the value of  sin 3  i cos 3  Solution De Moivre’s theorem requires this to be of the form (cos θ + i sin θ)n Hence  sin 3  i cos 3  must be changed either into  cos 6  i sin 6 

or into i  cos 3  i sin 3  which is i  cos   3   i sin   3   Thus  sin 3  i cos 3    cos 6  i sin 6  6

6

 cos   i sin   1

Example 4 Find the value of

1

3  3i 

3

Solution 3  3i  2 3  cos   6   isin   6  

1



3  3i  2 3  cos   3

1

 6   i sin   6  



3

3

3  1       cos   6   i sin   6   2 3 1   cos 2  i sin 2  24 3



1 3 i or i 72 24 3

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Edexcel FP2 Complex nos Sec. 2 Notes & Examples Using de Moivre’s theorem to find trigonometric identities The other application you study at this stage is the use of the theorem to give a multiple angle formula in terms of powers and to express powers of sine and cosine in terms of multiple angles. In the example below, the identities z n  z n z n  z n and cos n  are used frequently. You should be sin n  2i 2 happy with these identities and know how they arise.

Example 5 (i) Express sin6  in terms of multiple angles (ii) Hence find  sin 6  d .

putting n = 1

Solution (i)

z  cos  isin   sin n  So

 2i 

6

sin 6    z  z 1 

z n  z n  2isin   z  z 1 . 2i

6

.

 z 6  6 z 5 z 1  15 z 4 z 2  20 z 3 z 3  15 z 2 z 4  6 zz 5  z 6  z 6  6 z 4  15 z 2  20  15 z 2  6 z 4  z 6  z 6  z 6  6  z 4  z 4   15  z 2  z 2   20  2 cos 6  12 cos 4  30 cos 2  20 using z  z  2cos n for n = 6, 4 and 2 respectively n

Therefore: 

(ii)

 sin

6

n

64sin 6   2cos 6 12cos 4  30cos 2  20 . 20  2 cos 6  12 cos 4  30 cos 2 sin 6   64 10  cos 6  6 cos 4  15cos 2  32  10  cos 6  6 cos 4  15cos 2   d 32   1  1 6 15   10  sin 6  sin 4  sin 2   c 32  6 4 2  1   60  sin 6  9sin 4  45sin 2   c 192

 d   

Roots of unity First, think about the cube roots of unity. These are numbers such that z 3  1 or equivalently such that z 3  1  0 . Cast your mind back to the fundamental

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Edexcel FP2 Complex nos Sec. 2 Notes & Examples theorem of algebra: it tells you that the equation z 3  1  0 has three roots. One is obvious, it’s 1. This means that z  1 is a factor of z 3  1. Dividing z 3  1 by z  1 gives z 3  1  ( z  1)( z 2  z  1) . You can now find the other two roots of z 3  1  0 by applying the formula for roots of a quadratic to z 2  z  1  0 , as follows:

z

b  b2  4ac 1  1  4 1 3    i. 2a 2 2 2

The three roots of z 3  1  0 can be plotted on an Argand diagram.

1 3   i 2 2

Imaginary 0.8

2 3 2 3

-0.5

1 1

Real

-0.8

1 3   i 2 2 1 3 1 3 2 2 i is i is  The argument of   and the argument of   . 2 2 2 2 3 3 Notice that all the roots have modulus 1 and if you rotate any of the roots 2 through an angle of you obtain another one. Maybe you could have 3 guessed that these would be the three roots beforehand, bearing in mind De Moivre’s Theorem. This whole process can be repeated for fourth roots of unity as z 4  1  ( z  1)( z  1)( z 2  1) . The two complex fourth roots of unity are the roots of the quadratic which appears in this factorisation. Of course, the roots of z 2  1  0 are i so that the four fourth roots of unity are 1, 1,i, i . These four roots of z 4  1  0 can be plotted on an Argand diagram.

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Edexcel FP2 Complex nos Sec. 2 Notes & Examples

Imaginary 1

i

-1

11

-1

-1

Real

-i

Notice again that all the roots have modulus 1 and if you rotate a root through  2 an angle of  you obtain another one. The fifth and sixth roots of unity 2 4 look like this: Fifth roots Imaginary 1

of unity Imaginary 1

-1 -1

1

1

Real

Real

Sixth roots of unity -1 -1

In summary, the nth roots of unity can be expressed as

z  cos

2k 2k  i sin n n

for k = 0, 1, 2, …, n - 1

and the sum of all the nth roots of unity is always zero (make sure that you understand why this is the case).

General nth roots To find the nth root of any number, find one root and the others are placed symmetrically around a circle, centre 0, in the Argand diagram.

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Edexcel FP2 Complex nos Sec. 2 Notes & Examples Example 6 Find the four roots of z4 = –16. Convert –16 to reiθ form

Solution z 4  16ei

z  16  4  ei  1

1 4

i

 2e 4 On an Argand diagram, this point lies on a circle of radius 2. z4 = –16 has four roots. These are symmetrically placed round the circle of radius 2.

2 Remember that θ must lie between –π and π i

3i 4

 4



i 4



3i 4

Thus they are: 2e , 2e , 2e and 2e . Written in the x + iy form, these four roots are  2 (1  i)

Example 7 Find the six roots of (z – 2)6 = –27i

Solve the equation for z – 2 and then find z.

Solution ( z  2)  27e 6



i 2 1

 One root for z – 2 is (27) 6 (e



i 2

1

3e

)6 =



i 12

On an Argand diagram, this point lies on a circle of radius 3 . (z – 2)6 = –27i has six roots. These are symmetrically placed round the circle of radius 3 .

Remember that θ must lie between –π and π

Thus they are: z–2=

3e



i 12

11i

7i

i

,

3e 4 ,

3e 12 ,

Hence the six roots for z are 2  3e



2  3e

i 12



3e 12 ,

3e

i



5i 12

,

3e 7i



3i 4

11i

, 2  3e 4 , 2  3e 12 , 2  3e 12 ,

5i 12

, 2  3e



3i 4

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Edexcel FP2 Complex nos Sec. 2 Notes & Examples As for the roots of unity, the sum of all the nth roots of a complex number is always zero. Make sure that you understand why this is the case, both algebraically (using the sum of a geometric series) and geometrically (using vector addition).

You can explore the geometrical pattern formed by the roots of a complex number using the Flash resource Complex roots.

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