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.
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.
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