If $z$ is a complex number of unit modulus and argument $\theta$ then $\arg \left(\frac{z^{5}-\bar{z}}{z^{5}+\bar{z}}\right)$ is?

Question

Let $z$ be a complex number of unit modulus and argument $\theta$. Calculate $$\arg \left(\frac{z^{5}-\bar{z}}{z^{5}+\bar{z}}\right)$$

My approach: I just tried a very general approach. So basically $z\bar{z} = |{z}|^2$ and since its unit modulus I essentially wrote $\bar{z}$ as $\frac{1}{z}$ and tried solving it. It gives me a scenario where I have to basically find out $z^5$ or $z^6$ and then try doing it the long way. This certainly doesn't seem to me like the intended solution. I believe there must be some better way to do this which I am not able to figure out.

Any help on approach or hints would do! Thanks for your time!

Answer 1

Let $$w = \frac{z^{5}-\bar{z}}{z^{5}+\bar{z}}$$

We have $\bar{z}= {1\over z}$ since $|z|=1$, so $$w = \frac{z^{3}-\bar{z}^3}{z^{3}+\bar{z}^3} \implies \bar{w} = -w$$ which means that $w$ is on imaginary axis, and thus $\arg (w) = \pm {\pi\over 2}$

Answer 2

$z$ being with modulus $1$, it can be written $z=e^{i\theta}$. Therefore:

$$\frac{z^{5}-\bar{z}}{z^{5}+\bar{z}}=\frac{e^{5i \theta}-e^{-i \theta}}{e^{5i \theta}+e^{-i \theta}}=\frac{e^{2i \theta}(e^{3i \theta}-e^{-3i \theta})}{e^{2i \theta}(e^{3i \theta}+e^{-3i \theta})}=\frac{2i \sin(3 \theta)}{2 \cos(3 \theta)}=i \tan 3\theta$$

Can you find the solution from there ?

Answer 3

$$\frac{z^{5}-\bar{z}}{z^{5}+\bar{z}} = \frac{z^{5}-\bar{z}}{z^{5}+\bar{z}} \frac{\bar{z}^{5}+z}{\bar{z}^{5}+z} $$ $$= \frac{1-1+z^6-z^{-6}}{1+1+z^6+z^{-6}} $$ $$= \frac{2i \sin 6 \theta}{2+2 \cos 6 \theta} $$ which is a pure imaginary number and so its argument is $\pi/2$.

Answer 4

$$ \frac{z^{5}-\bar{z}}{z^{5}+\bar{z}} = \frac{z^{6}-1}{z^{6}+1} = \frac{w-1}{w+1} $$ where $w = z^6$ lies on the unit circle. The points $-1, w, +1$ form a triangle in the plane, and according to Thales's theorem, the angle at the point $w$ is a right angle.

It follows that $\arg \left(\frac{w-1}{w+1}\right)$ is $+\frac\pi 2$ or $-\frac\pi 2$, depending on whether $w$ lies in the upper or lower half-plane.

The argument is undefined if $w=z^6 = \pm 1$.