# roots of complex numbers

• May 13th 2008, 07:12 AM
samdmansam
roots of complex numbers
a) find all roots of the equation z^5=sqrt(3)*i-1, present your answer in polar form

b) express (2-2i)^5 in exact polar form
(-1+sqrt(3)*i)^4

I was alright until the whole polar form thing, your help would be greatly appreciated.
• May 13th 2008, 08:57 AM
topsquark
Quote:

Originally Posted by samdmansam
a) find all roots of the equation z^5=sqrt(3)*i-1, present your answer in polar form

For a) what is the polar form of $-1 + i\sqrt{3}$?

$r~cos(\theta) + ir~sin(\theta) = -1 + i\sqrt{3}$

So
$r~cos(\theta) = -1$
and
$r~sin(\theta) = \sqrt{3}$

Thus
$\frac{r~sin(\theta)}{r~cos(\theta)} = -\sqrt{3}$

Now,
$tan(\theta) = \left | -\sqrt{3} \right | \implies \theta = \frac{\pi}{3}$

and since sine is positive and cosine negative we know the reference angle is in QII. Thus
$\theta = \frac{2 \pi}{3}$

Then
$r~cos(\theta) = -1$

$r \cdot -\frac{1}{2} = -1$

$r = 2$

So
$z^5 = 2e^{2i \pi / 3}$

Can you solve it from here? (Please note that there are five answers to this.)

-Dan