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

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...

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1 Answer 1

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.

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Thanks Dr. Israel. It's all clear now. –  Tian He Sep 26 '12 at 1:14

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