# How do I plot $-\frac{\pi}{2} \leq \operatorname{Arg}\left(\frac{z}{\bar{z}}\right) \leq\frac{\pi}{2}$ on the complex plane? [duplicate]

I'm new to Complex numbers, so I don't know a lot about them yet.

How do I plot this condition on the complex plane?

$$-\frac{\pi}{2} \leq \operatorname{Arg}\left(\frac{z}{\bar{z}}\right) \leq\frac{\pi}{2}$$

• This is a duplicate question that I have just answered as suggested above – Henry Lee May 23 '18 at 15:35
• @Skatinima Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – user Jun 22 '18 at 20:14

HINT

Note that

$$\frac{z}{\bar{z}}=\frac{z^2}{\bar{z}z}=\frac{z^2}{|z|^2}\implies \operatorname{Arg}\left(\frac{z}{\bar{z}}\right)=2\operatorname{Arg}\left(z\right)$$

then we need to plot the region for $z$ such that

$$-\frac{\pi}{4} \leq \operatorname{Arg}\left(z\right)\leq\frac{\pi}{4}$$

plot of the region

Hint: If $z=\rho e^{i\theta}$, then $\dfrac{\,z\,}{\overline z}=e^{2i\theta}$.