# Thread: De'Moivre's theorem in polar form. Stuck

1. ## De'Moivre's theorem in polar form. Stuck

"Use De Moivre's theorem to solve the following equation, in polar form."

$\displaystyle z^{3}=i$

My working out:
Magnitude:
$\displaystyle z=1$
Angle?
$\displaystyle tan(\theta)=0$
I'm not sure what to do because it's 0 isn't it?

2. Angle is $\displaystyle \displaystyle\frac{\pi}{2}$

$\displaystyle (\cos{\frac{\pi}{2}+i\sin{\frac{\pi}{2})^3$

3. $\displaystyle z^{3}=cis(\frac{\pi}{2})$
$\displaystyle z=cis(\frac{\pi}{2} \times \frac{1}{3} )$
$\displaystyle z=cis(\frac{\pi}{6})$
Ah, I see. Thanks.

5. $\displaystyle \displaystyle e^{i (\frac{\pi}{6} + \frac{2 \pi}{3} n)}$ for n = 0,1,2.
$\displaystyle z^{3}=cis(\frac{\pi}{2})$
$\displaystyle z=cis(\frac{\pi}{2} \times \frac{1}{3} )$
$\displaystyle z=cis(\frac{\pi}{6})$
You have to remember that there are 3 cube roots, and they are evenly spaced around a circle. So they are separated by an angle of $\displaystyle \displaystyle \frac{2\pi}{3}$.