Let z be a non zero complex number; we can write z in the polar form as,
z = r(cos θ + i sin θ) = r e^{iθ}, where r is the modulus and θ is argument of z.

Important point
Multiplying a complex number z with e^{iα} gives
z *e^{iα} = re^{iθ} X e^{iα} = re^{i(α + θ)}
The resulting complex number re^{i(α + θ)} will have the same modulus r and argument (α + θ).

De Moivre's theorem

De Moivre's theorem states following cases

Case I
It states that for any integer n,

(cos θ + i sin θ)^{n} = cos (nθ) + i sin (nθ)

Case II
if n is of the form p/q where p, q are integers and q > 0
then
(cos θ + i sin θ)^{n} =cos (2k π+ θ)p/q + sin (2k π+ θ)p/q
Where k=0,1,2,...q-1

Proof for De Moivre's theorem(Case -1)

Prove using Euler's Formula

This can be easily proved using Euler's formula as shown below.

We know that, (cos θ + i sin θ) = e^{iθ}
(cos θ + i sin θ)^{n} = [e^{iθ}]^{n}=e^{i(nθ)}
Therefore,
e^{i(nθ)} = cos (nθ) + i sin (nθ)

Prove using Mathematical Induction ( n being a positive number)
(cos θ + i sin θ)^{n} = cos (nθ) + i sin (nθ)

Step 1 for n=1
(cos θ + i sin θ)^{n}=(cos θ + i sin θ)

Step 2 for n=m,let us assume the theorem is true for n=m
(cos θ + i sin θ)^{m} = cos (mθ) + i sin (mθ)

step 3Now for n=m+1
(cos θ + i sin θ)^{m+1}
=(cos θ + i sin θ)^{m}(cos θ + i sin θ)
=[cos (mθ) + i sin (mθ)](cos θ + i sin θ)
=[cos(mθ) cosθ -sin(mθ)sin θ] +i[cos (mθ)sin θ + sin (mθ)cos θ]
=cos (m+1)θ + i sin (m+1)θ
So, theorem holds good for n=m+1
So, by principle of Mathematical Induction

(cos θ + i sin θ)^{n} = cos (nθ) + i sin (nθ)

For negative integers, we can write like n=-m where m is positive number
(cos θ + i sin θ)^{n} =1 /(cos θ + i sin θ)^{m}

Now m being positive, we already proved above
=1 /cos (mθ) + i sin (mθ)
Multiplying the Conjugate complex on Numerator and Denominator we get,
=cos (mθ) - i sin (mθ)
=cos (-mθ) +i sin (-mθ)
=cos (nθ) + i sin (nθ)

Application of De Moivre's theorem

De Moivre's theorem is used to calculate the root of Complex Numbers

For a complex number z=a + ib which can be expressed as polar form as = r(cos θ + sin θ)
So z^{1/n} = r^{1/n}(cos θ + sin θ)^{1/n}
Now from De Moivre's theorem, we know that
(cos θ + i sin θ)^{1/n} =cos (2k π+ θ)/n + sin (2k π+ θ)/n
Where k=0,1,2,...n-1

By putting value of k=0,1,2,...n-1, we can find the root values