# 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?

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!

• $z=\cos\theta+i\sin\theta$, so $\arg(z^6\pm 1)=\arg[(\cos 6\theta\pm 1)+i\sin 6\theta]$, $\arg(\frac{z^5-\bar z}{z^5+\bar z})=\arg(z^6-1)-\arg(z^6+1)\pmod{2\pi}$ etc. May 1 at 14:43

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

So 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 $$\arg (w) = \pm {\pi\over 2}$$

$$z$$ 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 ?

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

• ... or $-\pi/2$. May 1 at 14:49

$$\frac{z^{5}-\bar{z}}{z^{5}+\bar{z}} = \frac{z^{6}-1}{z^{6}+1} = \frac{w-1}{w+1}$$ 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 $$w=z^6 = \pm 1$$.