There's one of these that I am completely stuck on.
The two equations referred to in the title are:
$\lvert Re(z-5)\rvert\leq1$ and $Re(\frac{z}{\bar{z}})=1$
The second equation will always hold true because complex conjugate does nothing to it's real part.
From the first equation, I conclude easily enough that the interval is $x \in [4,6]$.
But that leaves me with an infinite amount of complex numbers on that interval without a defined imaginary part, and I don't know how to sketch those out.
Write $\;z=a+ib\;,\;\;a,b\in\Bbb R\;$ , so:
$$z-5=(a-5)+ib\implies\left|\text{Re}\,(z-5)\right|=|a-5|\le1\iff 4\le a\le 6$$
and you have a pretty precise closed interval on the real axis.
Now,
$$\frac z{\overline z}=\frac{(a+ib)^2}{a^2+b^2}=\frac{a^2-b^2}{a^2+b^2}+i\frac{2ab}{a^2+b^2}\implies \text{Re}\,\frac z{\overline z}=\frac{a^2-b^2}{a^2+b^2}=1\implies\ldots\text{etc.}$$
Solve now.