## Euler's formula

- For any real number x, e
^{ix} = cos x + i sin x
- 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 3*Now 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

**Also Read**

**Notes**
**NCERT Solutions & Worksheets**