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Complex numbers - Complex Transformations 1

Submitted by roamingfree on Sat, 08/28/2021 - 17:28
skills
Apply mobius transformations to determine new loci
Question

Let $C$ be the locus of complex points defined by $|z| = 2$. Determine the image of $C$ under the transformation $$w = \frac{z+1}{z-1}$$

solution -- press button to toggle

To proceed, we first express $z$ in the form $z = f(w)$. $$\begin{align} w(z-1) &= z+1 \\ wz - w &= z+1 \\ wz - z &= w + 1 \\ z(w-1) &= w + 1\\ z &= \frac{w+1}{w-1} \end{align}$$

This rearrangement gives the loci of the transformed curve has equation $$ \left|\frac{w+1}{w-1}\right| = 2 $$ rearrangement of this yields $$ \left|w+1\right| = 2\left|w-1\right| $$ (note that this is a circle, since $|w - \alpha| = k|w-\beta|$ is a circle is $k\neq 1$ and a straight line in the case $k=1$

We now substitute the $w = u + iv$, this gives $$ \begin{align} \left|u+iv+1\right| &= 2\left|u+iv-1\right| \\ (u+1)^2 + v^2 &= 4\left((u-1)^2 +v^2\right) \\ u^2 + 2u + 1 + v^2 &= 4u^2 - 8u +4 + 4v^2 \\ 0 &= 3u^2 -10u + 3v^2 + 3 \\ 0 &= 3\left(u - \frac{5}{3} \right)^2 - \frac{25}{9} + 3 +3v^2 \\ 0 &=3\left(u - \frac{5}{3}\right)^2 +3v^2 +\frac{2}{3}\\ 0 &= \left(u - \frac{5}{3}\right)^2 + v^2 + \frac{2}{9}\end{align} $$

We conclude that the image of $C$ is a circle with centre $z= \frac{5}{3}$ and radius $\frac{\sqrt{2}}{3}$