Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$\Gamma = { B\over e^{j\theta} -A}$$

Both $A$ and $B$ are complex numbers.

The tedious way of course is to expand $A$, $B$ and $e^{j\theta}$, formulate the function into the form of $\Gamma = x + jy$, then prove $x^2 + y^2 = r^2$.

But I wonder whether there's a more clever way...

High school was such a long time ago and I find myself unable to come up with clever tricks any more...

share|cite|improve this question
up vote 3 down vote accepted

$z = e^{j\theta}$ for real $\theta$ traces out the unit circle. A linear fractional transformation such as $z \to B/(z - A)$ takes circles (and straight lines) to circles (or straight lines). If $A$ is on the unit circle the image is a straight line; if not, it is a circle.

EDIT: For convenience we may assume $B = 1$ (since scalings and rotations take circles to circles). Suppose $|A| \ne 1$. Let $p = \dfrac{\overline{A}}{1-|A|^2}$. Then $$ \frac{1}{e^{j\theta} - A} - p = \frac{1-\overline{A} e^{j\theta}}{(e^{j\theta}-A)(1-|A|^2)}$$ But $$\left|1 - \overline{A} e^{j\theta}\right| = \left|e^{j\theta} (e^{-j\theta} - \overline{A})\right| = \left|e^{j\theta}-A\right|$$ so $$\left| \frac{1}{e^{j\theta} - A} - p \right| = \frac{1}{|1-|A|^2|}$$ is constant. Thus $\dfrac{1}{e^{j\theta}-A}$ describes a circle with centre $p$ and radius $1/{|1 - |A|^2|}$.

I'll leave the case $|A|=1$ to you.

share|cite|improve this answer
Thanks Dr. Israel. It's all clear now. – Tian He Sep 26 '12 at 1:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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