Find image $f(z)=1/z$ given set $\{z=x+iy \in \mathbb{C} : \ (x-1)^2 + y^2 = a^2\}$

Question

Let $f(z)=1/z$. Consider the set $S = \{x+iy \in \mathbb{C} : \ (x-1)^2 + y^2 = a^2\}$. Find the image of $S$ under $f$. The hint is to consider $0 < a < 1$ separate from $a \geq 1$.

I'm not sure how to visualize this -- we haven't spent any time in lecture on this. My attempt: we can write $f(z) = 1/z = \overline{z}/|z|^2 = (x-iy)/(x^2+y^2)$. Obviously the image of $S$ under $\overline{z}$ is exactly the same circle since each point of the circle gets flipped about the $x$-axis. However, I'm not sure how the scaling $1/|z|^2$ affects things, since it depends on the point $z$.

Answer 1

The equation of a circle is $$z\bar z+a \bar z+\bar a z+c~~~~(1)$$ where $z=x+iy$, $a$ is complex and $c$ is real. let $w=1/z$ (inversion), where $w=u+iv$, then we get $$1+a \bar w+ \bar a w +c w \bar w=0, ~~~~(2)$$ which again is a circle in $w$-plane. If $c=0$, (2) will be a line.

The given equation of the circle $$(x-1)^2+y^2=a, 0< a <1$$ can also be written as $$[z-1|=a ~~~~(3)$$ so under inversion $z=1/w$, we get $$|1-w|=a|w| \implies (u-1)^2+v^2=a^2(u^2+v^2) $$
$$\implies u^2+v^2-\frac{2u}{1-a^2}+\frac{1}{1-a^2}=0$$ which is again a circle in $(u,v)$ plane. Finally, one can write this image-circle as $$x^2+y^2-\frac{2x}{1-a^2}+\frac{1}{1-a^2}=0 $$

Answer 2

Note that the circle $(x-1)^2 + y^2 = a^2$ is equivalent to the complex equation $|z-1|=a$. Let $w = f(z) = \frac1z$ and rewrite the equation $|z-1|^2=a^2$ as,

$$(\frac1w - 1)(\frac1{\bar w}-1) = a^2$$

Rearrange,

$$|w|^2 + \frac1{a^2-1}(w+\bar w)= \frac1{a^2-1}$$

or, in the familiar form of a circle,

$$\bigg|w + \frac1{a^2-1} \bigg|^2 = \frac {a^2}{(a^2-1)^2}$$

Thus, the image of $f(z)=w$ is a circle with center at $-\frac1{a^2-1}$ and radius $\frac{|a|}{|a^2-1|}$.