# 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

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:

$$\frac{z\overline{z}-a\overline{z}-\overline{a}z+a\overline{a}}{1-\overline{a}z-a\overline{z}+a\overline{a}z\overline{z}}=\frac{(z-a)(\overline{z}-\overline{a})}{(1-\overline{a}z)(1-a\overline{z})}=\frac{|z-a|^2}{|1-\overline{a}z|^2}<1$$

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 $|a|=1$.

• so you're saying that this is a circle with radius |a-bar| and has to end points, a and 1/a? Interesting. Is there no way to simplify the equation down a little bit more to actually see that it is a circle? You know, like getting x^2 + y^2 = ..... or something like that? Either way, thanks. – Joe Feb 2 '15 at 18:30
• @Joe Yes, it can be done. This is just one way, which avoids getting your hands too dirty. – Pp.. Feb 2 '15 at 18:31
• Haha, fair enough. I just want to be able to expand it out so I can see how this "a" parameter works. – Joe Feb 2 '15 at 18:35
• @Joe See the second option. – Pp.. Feb 2 '15 at 18:38
• many thanks man, huge help!! – Joe Feb 2 '15 at 18:41

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 $C = \{ z | \operatorname{re} ((\overline{a-b}) (z- {a+b \over 2})) = 0 \}$.)

• oh man, I didn't see this until now. thanks for your response! – Joe Feb 6 '15 at 0:39
• Glad to be able to help! (This is meant as an addition to PPs. answer.) – copper.hat Feb 6 '15 at 1:02