Determining $\arg(u-\sqrt{3})$ in exact form. Let $u$ be the solution to $z^5=-9\sqrt{3}i$ so that $\frac{\pi}{2}\le arg(u) \le \pi$.
Determine $\arg(u-\sqrt{3})$ in exact form.
How would I go about completing this question?
 A: Let $z=r(\cos\theta+i\sin\theta)$. We know $-9i\sqrt3=9\sqrt3(\cos-\frac{\pi}{2}+i\sin-\frac{\pi}{2} )$
This means (using De Moivre's Theorem)
$$r^{5}(\cos5\theta+i\sin5\theta)=9\sqrt3(\cos-\frac{\pi}{2}+i\sin-\frac{\pi}{2})=9\sqrt3(\cos(-\frac{\pi}{2}+2k\pi)+i\sin(-\frac{\pi}{2}+2k\pi)$$
So $r^{5}=9\sqrt3$ so $r=\sqrt3$.
We also have $5\theta=-\frac{\pi}{2}+2k\pi$ so $\theta=-\frac{\pi}{10}+\frac{2k\pi}{5}$
For different values of $k$:
$$k=0: \theta=-\frac{\pi}{10}$$
$$k=1: \theta=-\frac{\pi}{10}+\frac{2\pi}{5}=\frac{3\pi}{10}$$
$$k=-1: \theta=-\frac{\pi}{10}-\frac{2\pi}{5}=-\frac{\pi}{2}$$
$$k=2: \theta=-\frac{\pi}{10}+\frac{4\pi}{5}=\frac{7\pi}{10}$$
$$k=-2: \theta=-\frac{\pi}{10}-\frac{4\pi}{5}=-\frac{9\pi}{10}$$
Ignore all values of $\theta<\frac{\pi}{2}$ as this is what you specified. This leaves us with : $$\theta= \frac{7\pi}{10}$$
So (using Euler's relation, that $e^{i\theta}=\cos\theta+i\sin\theta)$:
$$ u=\sqrt3 e^{\frac{7i\pi}{10}} $$
Can you do it from there?
Hope that helped :)
A: Green points represent solutions of the equation $\;z^5=-9\sqrt{3}i.$
From $|z^5|=(\sqrt3)^5\;$ it follows that $|u|=\sqrt3,\,$ and the points with affices $\;0,u,u-\sqrt3,-\sqrt3\;$ are vertices of a rhombus. Therefore, diagonals are also angle bissectors. We have
$$\arg(u-\sqrt3)=\frac{7\pi}{10}+\beta,$$
where $$\frac{7\pi}{10}+2\beta=\pi.$$
This gives
$$\arg(u-\sqrt3)=\frac{17\pi}{20}.$$

