# Showing that the map $f(z) = \frac{1}{z}$ maps circles into circles or lines

Let $f: \mathbb{C} \setminus \{0 \} \to \mathbb{C} \setminus \{0\}$. We want to show that $f(z) = \frac{1}{z}$ maps circles into circles and lines. My professor gave the following hint: The general equation for lines and circles is

$$\alpha(x^2 + y^2) + \beta x + \gamma y + \Delta = 0$$

where the greek letters are obviously constants. So, given this advice, We can rewrite this in the complex plane as follows:

$$\alpha |z|^2 + \frac{ \beta}{2}( z + \overline{z} ) + \frac{\gamma}{2i}( z - \overline{z} ) + \Delta = 0$$

So, now we apply $w = \frac{1}{z}$ and we obtain (with $|w|^2 = w \overline{w}$):

$$\frac{ \alpha}{w \overline{w}} + \frac{\beta}{2}\bigg( \frac{1}{w} + \frac{1}{\overline{w}}\bigg) + \frac{\gamma}{2 i}\bigg( \frac{1}{w}- \frac{1}{\overline{w}}\bigg)+ \Delta = 0$$

hence,

$$\alpha + \frac{ \beta}{2}(\overline{w} + w ) + \frac{\gamma}{2i}(\overline{w}-w) + w \overline{w} \Delta = 0$$

Next, putting $w = u + iv$ we arrive to:

$$\alpha + \beta u - \gamma v + (u^2 + v^2 ) \Delta = 0$$

So, in the case when we have circles in $xy$-plane, that is when $\alpha \neq 0$, we still have circles in the $uv$-plane. So $f$ sends circles to circles if $\alpha \neq 0$. We also have circle if $\alpha \neq 0$ and $\Delta = 0$ which in this case $\frac{1}{z}$ sends circles to lines.

Is this a correct solution? IS there a shorter way to prove this?

• Looks OK to me. I'd add that a circle gets mapped to a line if and only if the circle passes through $0$, and a line gets mapped to a circle if and only if the line does not pass through $0$. Commented Dec 13, 2014 at 5:03
• Actually it's good to know that $z \mapsto \frac{1}{\overline{z}}$ is inversion, and inversion preserves (general) circles. This can be shown by elementary geometry and sometimes it's helpful to understand the geometric behaviour of the map. Commented Dec 13, 2014 at 5:17
• @MichaelHardy do you mean the circle gets mapped a line iff the circle is centered at $0$ ?
– user195835
Commented Dec 13, 2014 at 5:22
• No, circles that go through 0 become lines not passing through 0 and vice versa. Circles centred at 0 will remain centred at 0 but with possibly different radii. Lines through 0 will remain the same under inversion, but become reflected under the map you are considering. As I said, all these will be obvious if you understand inversion. See en.wikipedia.org/wiki/Inversive_geometry for the definition of inversion. Commented Dec 13, 2014 at 5:28
• No: circles centered at $0$ get mapped to other circles centered at $0$, or, in case the radius is $1$, the same circle centered at $0$. It is when $0$ is actually a point on the circle that the circle is mapped to a straight line. The reason is that the mapping $z\mapsto1/z$ takes $0$ to $\infty$ and the only "circles" that pass through $\infty$ are straight lines. ${}\qquad{}$ Commented Dec 13, 2014 at 5:57