Determining $\arg(u-\sqrt{3})$ in exact form.

Question

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?

Answer 1

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 :)

Answer 2

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 $$\arg(u-\sqrt3)=\frac{17\pi}{20}.$$