Backward Complex Transformation

Question

We need to show that transformation $w=\frac{1}{z}$ i.e. substituting $z=\frac{1}{w}$ map circles $|z-a|=a$ and $|z-ia|=a$ to vertical and horizontal line respectively.

Answer 1

Let $w = u+iv$

We have $ z = \frac1w$ $$|z-a| = a \Rightarrow \left\vert\frac{1-aw}{w}\right\vert = a \Rightarrow |1-a(u+iv)| = a|u+iv|$$

$$ (1-au)^2+a^2v^2 = a^2u^2 + a^2v^2 \Rightarrow 1- 2au = 0 \Rightarrow \color{blue}{u = \frac{1}{2a}}$$

This represents a straight line

Can you do for the second one now?

Answer 2

Here is a different way : Let us take $z$ belonging to the first circle. Squaring, we have :

$$|z-a|^2=a^2 \ \iff \ (z-a)(\bar z -a)=a^2 \ \iff \ |z|^2-a(z+\bar z)+a^2=a^2 \ \iff $$

$$\ |z|^2=a(z+\bar z) \tag{1}$$

Now consider the set of points of the form $1/z$. We can write, using (1) and $z=x+iy$ :

$$\dfrac 1 z = \dfrac{\bar z}{z \bar z}= \dfrac{\bar z}{|z|^2}= \dfrac{\bar z}{a (z+\bar z)}=\dfrac 1 a \dfrac{x-iy}{2x}=\dfrac 1 a (\dfrac{1}{2}-i\dfrac{y}{x})$$

i.e., the locus is the set of points whose real part is $\dfrac{1}{2a}$ and whose imaginary part can be ... anything.

The answer is : the vertical line with cartesian equation $x=\dfrac{1}{2a}$.