Plotting Reciprocal of a Complex inequality We have a complex inequality $|z-r|\leq{\sqrt{r}}$. (${r>0}$ is a real constant number and $z$ is a variable complex number). Now we want to find the plot of $f(z) = 1/z$ where $z$ are the answers of above inequality. i.e. we want to find plot of: $|1/z-r|\leq \sqrt{r}$ $~~~$
I know that the first one is the plot of a circle with Radius $=\sqrt r$ and center is $r + 0 i$. (All the point inside and on the perimeter of circle).
But I don't know what to do with the reciprocal  of $z$ in $|1/z-r|\leq \sqrt{r}$ .
 A: Let $D$ be the unit disc, then your original inequality is satisfied by is the set
$$
r + z\sqrt{r}, \quad z \in D
$$
and the inverse would be
$$
\frac{1}{r+z\sqrt{r}}, \quad z \in D.
$$
A: Recall that if a point in the plane is represented by $z=x+iy$ for $x,y\in \mathbb{R}$, then you have
$$
\frac{1}{z}=\frac{\bar{z}}{|z|^2}
\quad \implies \quad
\frac{1}{z-z_0}=\frac{\overline{z-z_0}}{|z-z_0|^2}.
$$
Take $z_0=x_0+iy_0$. This tells you that if you have $|z-z_0|\le R$ for real number $R>0$, then $\frac{1}{z-z_0}$ yields another circle for $\overline{z-z_0}$ scaled by $|z-z_0|^2\le R^2$. This is can be seen from
$$
\frac{1}{z-z_0}
=\frac{\bar{z}-\bar{z_0}}{|z-z_0|^2}
\ge \frac{\bar{z}-\bar{z_0}}{R^2}.
$$ 
In your case, you have 
$$
\left|\frac{1}{z-r}\right|
=\left|\frac{\bar{z}-r}{|z-r|^2}\right|
=\frac{|\bar{z}-r|}{|z-r|^2}
\ge \frac{|\bar{z}-r|}{r}
=\frac{\sqrt{r}}{r}
=\frac{1}{\sqrt{r}}.
$$
What you want is the plot of the circle described by $\left|\frac{\bar{z}-r}{r}\right|\le \frac{1}{\sqrt{r}}.$ That's it. But then you have to take into account the change of angle $\theta$ when you write $z=\sqrt{r}e^{i\theta}$ then $\bar{z}=\sqrt{r}e^{-i\theta}$. 
