De Moivres Theorem

De Moivre's theorem can be proved using mathematical induction.

Consider the simplest case of De Moivre's theorem, when

n = 1 .

This can be represented with the following equation.

[r (cos θ + i sin θ)]^{1} = r^{1} (cos 1 θ + i sin 1 θ)

Both sides of this equation simplify to

r (cos θ + i sin θ)

because multiplying a number by

1

or raising it to the power of

1

does not change its value.

[r (cos θ + i sin θ)]^{1} r (cos θ + i sin θ) = r^{1} (cos 1 θ + i sin 1 θ) ⇓ = r (cos θ + i sin θ)

Since both sides are identical, the base case holds true.

Assume that the statement is true for some natural number

k .

This assumption implies that the following equation is true.

z^{k} = r^{k} (cos k θ + i sin k θ)

This assumption serves as the basis for the inductive step.

It needs to be shown that a complex number raised to the power of

k + 1

is another complex number with the same modulus raised to the power of

k + 1

and the argument multiplied by

k + 1 .

z^{k + 1} = r^{k + 1} (cos (k + 1) θ + i sin (k + 1) θ)

To establish this, expand the power to be the product of the complex number raised to the power of

k

and the number.

z^{k + 1} = z^{k} \cdot z

The expression can be expanded further by using the inductive hypothesis.

z^{k + 1} = [r^{k} (cos k θ + i sin k θ)] [r (cos θ + i sin θ)]

Next, multiply the complex numbers and apply trigonometric identities to simplify the expression.

[r^{k} (cos k θ + i sin k θ)] [r (cos θ + i sin θ)]

r^{k + 1} (cos k θ \cdot cos θ + i cos k θ \cdot sin θ + i sin k θ \cdot cos θ + i^{2} sin k θ \cdot sin θ)

$i^{2} = - 1$

r^{k + 1} (cos k θ \cdot cos θ + i cos k θ \cdot sin θ + i sin k θ \cdot cos θ - sin k θ \cdot sin θ)

Commutative Property of Addition

r^{k + 1} (cos k θ \cdot cos θ - sin k θ \cdot sin θ + i cos k θ \cdot sin θ + i sin k θ \cdot cos θ)

Factor out $i$

r^{k + 1} [cos k θ \cdot cos θ - sin k θ \cdot sin θ + i (cos k θ \cdot sin θ + sin k θ \cdot cos θ)]

$cos (α) cos (β) - sin (α) sin (β) = cos (α + β)$

r^{k + 1} [cos (k θ + θ) + i (cos k θ \cdot sin θ + sin k θ \cdot cos θ)]

$sin (α) cos (β) + cos (α) sin (β) = sin (α + β)$

r^{k + 1} (cos (k θ + θ) + i (sin (k θ + θ))

Factor out $θ$

r^{k + 1} (cos (k + 1) θ + i (sin (k + 1) θ)

This result matches the right-hand side of De Moivre's theorem for

n = k + 1,

confirming that the theorem holds for all positive integers

n .

Therefore, by mathematical induction, De Moivre's theorem is proven to be true.

$z^{n} = r^{n} (cos n θ + i sin n θ)$

De Moivre's Theorem