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

b) express(2-2i)^5in exact polar form

(-1+sqrt(3)*i)^4

I was alright until the whole polar form thing, your help would be greatly appreciated.

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- May 13th 2008, 07:12 AMsamdmansamroots 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 AMtopsquark
For a) what is the polar form of $\displaystyle -1 + i\sqrt{3}$?

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

So

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

and

$\displaystyle r~sin(\theta) = \sqrt{3}$

Thus

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

Now,

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

and since sine is positive and cosine negative we know the reference angle is in QII. Thus

$\displaystyle \theta = \frac{2 \pi}{3}$

Then

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

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

$\displaystyle r = 2$

So

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

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

-Dan