Sketching an inequality in the complex plane so I have to sketch this inequality on the complex plane,
$$\frac {|z-a|} {|1- \bar az|}<1$$
where $|a| < 1$ is a complex number.
I know typically when there is just $z$'s and $i$'s  you replace $z$ with $z=x+iy$ and go from there (squaring both sides, using the modulus, etc..) but this "$a$" is throwing me off on what to do. Any tips please? 
Thanks
 A: One possible way: 
You inequality is (for $a\neq0$) equivalent to 
$$\frac{|z-a|}{\left|\frac{1}{\overline{a}}-z\right|}<|\overline{a}|$$
The equality 
$$\frac{|z-a|}{\left|\frac{1}{\overline{a}}-z\right|}=|\overline{a}|$$
says that the proportion of the distances from $z$ to the pair of points $a$ and $\frac{1}{\overline{a}}$ is a constant $|\overline{a}|$.
Apollonius theorem says this is a circle. The inequality is then one of the sides.

Another way: 
Your inequality is equivalent to 
$$\frac{z\overline{z}-a\overline{z}-\overline{a}z+a\overline{a}}{1-\overline{a}z-a\overline{z}+a\overline{a}z\overline{z}}=\frac{(z-a)(\overline{z}-\overline{a})}{(1-\overline{a}z)(1-a\overline{z})}=\frac{|z-a|^2}{|1-\overline{a}z|^2}<1$$
i.e. 
$$z\overline{z}-a\overline{z}-\overline{a}z+a\overline{a}<1-\overline{a}z-a\overline{z}+a\overline{a}z\overline{z}$$
or $$(1-a\overline{a})z\overline{z}<(1-a\overline{a}).$$
Depending on whether $|a|>1$ or $|a|<1$ we can cancel the $1-a\overline{a}$ and reverse or not the sign in the inequality.
We get $$|z|^2=z\overline{z}>1\text{ or }<1$$
Consider also the case $|a|=1$.
A: This is a slight elaboration of Pps. answer.
As noted in Pps. answer, we can write $\left| {z-a \over 1-z\bar{a}}\right |
= {1 \over |a|} {|z-a| \over |z-{1 \over \bar{a}}|}$, so we can
generalise slightly and consider the set of points  $C = \{z | \left| {z-a \over z-b}\right | = \lambda \}$, with $\lambda >0$ and $a \neq b$.
If we let $r e^{i \theta} = {b-a \over 2}$, and $w = e^{-i \theta}(z-{a+b \over 2})$, we obtain the equality
$\left| {w+r \over w-r}\right | = \lambda$, with $r \in \mathbb{R}$. Note that 
if $w$ satisfies the inequality, then so does $\bar{w}$, so the locus is symmetric about the real axis. Let $S = \{ w | \left| {w+r \over w-r}\right | = \lambda \}$.
If $\lambda = 1$, then by writing $|w+r|^2 = |w-r|^2$, it is straightforward to check that
$w \in S$ iff $\operatorname{re} w = 0$, that is $S$ is the imaginary axis.
If $\lambda \neq 1$, we have $w \in S$ iff $|w+r|^2 = \lambda^2|w-r|^2$,
or equivalently,
$|w|^2+r^2 + { 2(1+ \lambda^2) \over 1-\lambda^2} r \operatorname{re} w = 0$.
Note that if $c$ is real then $w$ solves the equation $|w-c| = \rho$ iff 
$|w|^2+c^2-\rho^2 -2 c\operatorname{re} w = 0  $. Comparing, we set
$c = -{ 1+ \lambda^2 \over 1-\lambda^2} r$ and
$\rho = \sqrt{c^2-r^2} = 2r { |\lambda|\over|1-\lambda^2| }$,
hence
$w \in S$ iff $|w+{ 1+ \lambda^2 \over 1-\lambda^2} r| = 2r { |\lambda|\over|1-\lambda^2| }$.
Now using $r={1\over 2}|b-a|$ and $e^{i \theta} = {b-a\over |b-a|}$, we can map this back to 'z' to get
$C= \{z | |z-z_0| = \rho \}$, where
$\rho = 2{ |\lambda|\over|1-\lambda^2| }(b-a)$ and
$z_0 = {a+b \over 2} + { \lambda^2+1 \over \lambda^2-1}{b-a \over 2} $.
In terms of the original problem, we have
$\lambda = |a|$, and $b= {1 \over \bar{a}}$, substituting these values
gives $\rho = 1, z_0 = 0$.
(And if $|a|=1$, the above gives $C = \{ z | \operatorname{re} ((\overline{a-b}) (z- {a+b \over 2})) = 0 \}$.)
