# What to say about (sq root 3 - i)^n = 2^n(cos n*(pie)/6 - i*sin n(pie)/6)?

### 1 Answer | Add Yours

You need to test if the given expression represents an identity, hence, you may start by converting the algebraic form of complex number `z = sqrt3 - i` into polar form such that:

`z = x + i*y`

`z = |z|(cos alpha + i*sin alpha)`

`{(|z| = sqrt(x^2 + y^2)),(alpha = tan^(-1)(y/x)):}`

Identifying `x = sqrt3` and `y = -1` yields:

`{(|z| = sqrt((sqrt 3)^2 + (-1)^2)),(alpha = tan^(-1)(-1/sqrt 3)):}`

`{(|z| = sqrt(4)),(alpha = -pi/6):}`

Replacing `2` for `|z|` and -`pi/6` for `alpha` yields:

`z = 2(cos (-pi/6) + i*sin (-pi/6))`

Using the identities `cos(-alpha) = cos alpha` and `sin(-alpha) = -sin alpha` yields:

`z = 2(cos (pi/6) - i*sin (pi/6))`

Raising to power n yields:

`z^n = 2^n(cos (pi/6) - i*sin (pi/6))^n`

Using De Moivre's theorem yields:

`z^n = 2^n(cos (npi/6) - i*sin (npi/6))`

Since `z = sqrt3 - i,` then `z^n = (sqrt3 - i)^n = 2^n(cos (npi/6) - i*sin (npi/6)).`

**Hence, testing if `(sqrt3 - i)^n = 2^n(cos (npi/6) - i*sin (npi/6))` holds, the answer is affirmative.**

**Sources:**