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 eiθ, where r is the modulus and θ is argument of z.
Important point
Multiplying a complex number z with eiα gives
z *eiα = reiθ X eiα = rei(α + θ)
The resulting complex number rei(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) = eiθ
(cos θ + i sin θ)n = [eiθ]n=ei(nθ)
Therefore,
ei(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 z1/n = r1/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
