# 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

- If z is a complex number, and n is a positive integer, then z
^{n}is a unique complex number. However z^{1/n}takes n distinct complex values. Thus a complex number has two square roots, three cube roots,..., n nth roots etc. - (cis )(cis ) = cis ( +) and cis /cis = cis ( -).
- (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 . - (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

z^{1/n} = r^{1/n}

Note that r^{1/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,

1^{1/3} = (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 +w +w² = 1 + (-1 +3 i)/2 + (-1 -3 i)/2 = 0
- 1. w. w² = , thus w³ = 1.
- 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.
- 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 z^{6}.

### 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,

z^{6} =

= 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 )³

### 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 x^{4} -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 x^{4} -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,

1^{1/4} = (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 -w^{10})(2 -w^{11}) = 49.

### Solution

(i) (1 +w -w²)^{6} = (-w² -w²)^{6}
[using 1 +w +w² = 0]

= (-2w²)^{6} = 2^{6}
w^{12} = 64(w³)^{4} = 64 [using w³ = 1]

(ii) (1 -w +w²)^{7} +(1 +w -w²)^{7} = (-w -w)^{7} +(-w² -w²)^{7}

= -2^{7}. w^{7} -2^{7} w^{14} = -2^{7}[(w³)².
w +(w³)^{4}. w²]

= -2^{7}. (w +w²) = -2^{7}. (-1) = 128.

(iii) (2 -w)(2 -w²)(2 -w^{10})(2 -w^{11})

= [4 -2(w +w²) +w³] [4 -2w^{10}(1 +w) +w^{21}]

= [4 -2(-1) +1] [4 -2w (-w²) +1]
(why?)

= (4 +2 +1)(4 +2. 1 +1) = 7. 7 = 49.

## Exercise

- Express each of the following in standard form x +iy

(i) (cos +i sin)^{10}(cos 2 +i sin 2 )^{4}

(ii) - Solve the equation x
^{4}-4 x² +8 x +35 = 0, given that one root is 2 + -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). - Prove that the complex cube roots of unity are reciprocal of each other.
- If 1, w and w² are cube roots of unity, show that

(i) (1 +w)(1 +w²)(1 +w^{4})(1 +w^{8}) = 1

(ii) (1 +w)(1 +w²)(1 +w^{4})(1 +w^{8})... upto 2n factors = 1

(iii) (1 -w)(1 -w²)(1 -w^{4})(1 -w^{8}) = 9 - Prove that = w or w²
- 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]