De-Moivre's Theorem (for rational index)

The theorem is stated in two steps as follows:
(i) (cos +i sin )n = cos n +i sin n , if n is an integer (positive, zero or negative)
(ii) cos n +i sin n is one of the values of (cos + i sin )n if n is a non integral rational number.

Remarks

  1. If z is a complex number, and n is a positive integer, then zn is a unique complex number. However z1/n takes n distinct complex values. Thus a complex number has two square roots, three cube roots,..., n nth roots etc.
  2. (cis )(cis ) = cis ( +) and cis /cis = cis ( -).
  3. (cos +i sin )-n = cos (-n ) +i sin (-n ) = cos n -i sin n .
    Thus, if z = cos +i sin then   1/z = cos -i sin .
  4. (cos -i sin ) n = (cos(- ) +i sin(- ))n = cos (-n ) +i sin (-n ) = cos n -i sin n .

Roots of a complex number

Let z = r cos = r cos ( +2 m ), where m I
       z1/n = r1/n
Note that r1/n is a real, positive number. Putting m = 0, 1, 2, ... n -1 we get n distinct roots of z, and then the values of start repeating.

Cube roots of unity

Since 1 = 1 +0 i = cis 0 = cis (0 +2m) = cis 2m, m I,
11/3 = (cis 2m)1/3 = cis 2m/3 where m = 0, 1, 2
= cis 0, cis 2/3, cis 4/3
     
= 1, (-1 +3 i)/2, (-1 -3 i)/2
Putting w = (-1 +3 1)/2
  
= (1 -3 -23 i)/4 = (-1 -3 i)/2
Hence three cube roots of unity are 1, w, w² where w = (-1 +3 i)/2

Remarks

  1. 1 +w +w² = 1 + (-1 +3 i)/2 + (-1 -3 i)/2 = 0
  2. 1. w. w² = , thus w³ = 1.
  3. It is easy to find cube roots of "a" in term of w, if a is a real number. Thus cube roots of 27 are 3, 3w, 3w²; cube roots of -8 are -2, -2w, -2w² etc.
  4. Note that 1, w, w² are vertices of an equilateral triangle in complex plane.

Illustrative Examples

Example

If z = (13 -5 i)/(4 -9 i), find z6.

Solution

Given z = (13 -5 i)/(4 -9 i) =[(13 -5 i)(4 +9 i)]/[(4 -9i)(4 +9i)]
= (97 +97 i)/97
= 1 + i/4 = 2 [cos(/4) + i sin (/4)]
Using Demoivre theorem,
z6 =
= 8(cos 3/2 +i sin 3/2) = 8[0 +i(-1)] = -8i

Example

Convert the following number to standard form x +iy
    (cos +i sin )7 (cos -i sin

Solution

(cos +i sin )7 (cos -i sin )³ = (cis )7 (cis (- ))³
     = (cis )7 ((cis )-1)³ = (cis )7 (cis )-3
     = (cis )4 = cis 4 = cos 4 +i sin 4 .

Example

Given that one of the roots of x4 -2 x³ +3 x² -2 x +2 = 0 is 1 +i, find the other three roots.

Solution

Since all the coefficients of the given equation are real numbers, the roots, if complex, occur in conjugate pairs. Thus two roots are 1 +i, 1 -i.
Hence x -(1 +i), x -(1 -i) are two factors of given equation,
i.e. (x -(1 +i))(x -(1 -i)) = x² -2 x +2 is a factor of given equation.
Now it is easy to see that x4 -2 x³ +3 x² -2 x +2 = (x² -2x +2)(x² +1).
Also we know that x² +1 = (x -i)(x +i).
Thus the four roots of the given equation are ±i, 1 ±i.

Example

Find the fourth roots of unity.
              

Solution

1 = 1 +0 i = cis 0
   = cis (0 +2 m) = cis 2m,        where m I,
11/4 = (cis 2 m )1/4 = cis 2m/4, where m = 0, 1, 2, 3
       = cis 0, cis /2, cis , cis 3/2
       = 1, i, -1, -i.
Hence the fourth roots of unity are ±1, ±i.

Example

Show that
(i) (1 +w -w²)6 = 64.
(ii) (1 -w +w²)7 +(1 +w -w²)7 = 128.
(iii) (2 -w)(2 -w²)(2 -w10)(2 -w11) = 49.

Solution

(i) (1 +w -w²)6 = (-w² -w²)6          [using 1 +w +w² = 0]
         = (-2w²)6 = 26 w12 = 64(w³)4 = 64 [using w³ = 1]
(ii) (1 -w +w²)7 +(1 +w -w²)7 = (-w -w)7 +(-w² -w²)7
    = -27. w7 -27 w14 = -27[(w³)². w +(w³)4. w²]
    = -27. (w +w²) = -27. (-1) = 128.
(iii) (2 -w)(2 -w²)(2 -w10)(2 -w11)
    = [4 -2(w +w²) +w³] [4 -2w10(1 +w) +w21]
    = [4 -2(-1) +1] [4 -2w (-w²) +1]                     (why?)
    = (4 +2 +1)(4 +2. 1 +1) = 7. 7  = 49.

Exercise

  1. Express each of the following in standard form x +iy
    (i) (cos +i sin) 10(cos 2 +i sin 2 )4
    (ii)  
  2. Solve the equation x4 -4 x² +8 x +35 = 0, given that one root is 2 + -3.
  3. Find
    (i) (4i)        (ii) -i         (iii) cube roots of -1
    (iv)             (v) cube roots of 64.
    Also find the product of roots in case of (iv) and (v).
  4. Prove that the complex cube roots of unity are reciprocal of each other.
  5. If 1, w and w² are cube roots of unity, show that
    (i) (1 +w)(1 +w²)(1 +w4)(1 +w8) = 1
    (ii) (1 +w)(1 +w²)(1 +w4)(1 +w8)... upto 2n factors = 1
    (iii) (1 -w)(1 -w²)(1 -w4)(1 -w8) = 9
  6. Prove that = w or w²
  7. Prove that is equal to
    (i) 2 if n is a multiple of 3
    (ii) -1 if n is not a multiple of 3.

Answers

1. (i) cos 2 +i sin 2
   (ii) cos /12 -i sin /12
2. Roots are 2 ±3 i, -2 ±i.
3. (i) ±2 (1 +i)       (ii) ±(1 -i)/2
   (iii) -1, (1 ± 3 i)/2; that is -1, -w, -w²
   (iv) ± (1 ±i)/2; product of roots is 1
    (v) 4, 4 w, 4 w², where w = (-1+3 i)/2; product of roots is 64.
6. [Hint. Let = x; then [-1-x] = x]