Sketching complex plane Sketch all points in the complex plane such that $Re(\frac{1}{z})=1$
I managed to solve it $(\frac{1}{z})= \frac{|z|}{z|z|}=\frac{(x - iy)}{(x^2+y^2)}$
Meaning that $Re(\frac{1}{z})= \frac{x}{(x^2+y^2)}$
How should I go afterwards?
 A: Nice start! Afterwards, you do\begin{align}\operatorname{Re}\left(\frac1z\right)=1&\iff\frac x{x^2+y^2}=1\\&\iff x=x^2+y^2.\end{align}If $x\notin[0,1]$, this equation has not solutions, because then $x<x^2\leqslant x^2+y^2$. On the other hand, if $x\in[0,1]$, you can take $y=\pm\sqrt{x-x^2}$.
So, the solution is$$\left\{x\pm\sqrt{x-x^2}\,i\,\middle|\,x\in(0,1]\right\}.$$Geometrically, this is the circle centered at $\frac12$ and with radius $\frac12$ (minus $0$).
A: For an alternative approach, use that $\,\operatorname{Re}(w)= \dfrac{1}{2}\left(w + \bar w\right)\,$ for complex $\,w\,$, then (assuming $\,z \ne 0\,$, otherwise $\,\dfrac{1}{z}\,$ is not even defined):
$$
\begin{align}
1=\operatorname{Re}\left(\frac{1}{z}\right)=\frac{1}{2}\left(\frac{1}{z}+\frac{1}{\bar z}\right) = \frac{z+\bar z}{2 z \bar z } \;\;&\iff\;\; z \bar z - \frac{z}{2} - \frac{\bar z}{2} \color{red}{+\frac{1}{4}-\frac{1}{4}} = 0 \\[5px]
 &\iff\;\; \left(z-\frac{1}{2}\right)\left(\bar z-\frac{1}{2}\right) = \frac{1}{4} \\[5px]
 &\iff\;\; \left|z - \frac{1}{2}\right|^2=\frac{1}{4} \\[5px]
 &\iff\;\; \left|z - \frac{1}{2}\right| = \frac{1}{2}
\end{align}
$$
The latter defines the set of points at constant distance $\,\dfrac{1}{2}\,$ from fixed point $\,\dfrac{1}{2}\,$, which is of course the circle of radius $\,\dfrac{1}{2}\,$ centered at $\,\dfrac{1}{2}\,$. However, the origin is excluded because $\,z \ne 0\,$, so the end answer is the same circle but "punctured" at $\,0\,$.
