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?



One possible way:

You inequality is (for $a\neq0$) equivalent to


The equality


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




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 $|a|=1$.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Joe Feb 2 '15 at 18:30
  • $\begingroup$ @Joe Yes, it can be done. This is just one way, which avoids getting your hands too dirty. $\endgroup$ – Pp.. Feb 2 '15 at 18:31
  • $\begingroup$ Haha, fair enough. I just want to be able to expand it out so I can see how this "a" parameter works. $\endgroup$ – Joe Feb 2 '15 at 18:35
  • $\begingroup$ @Joe See the second option. $\endgroup$ – Pp.. Feb 2 '15 at 18:38
  • $\begingroup$ many thanks man, huge help!! $\endgroup$ – 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 $C = \{ z | \operatorname{re} ((\overline{a-b}) (z- {a+b \over 2})) = 0 \}$.)

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.