# The Transformation $w=1/z$

Consider the equation $$w=\frac{1}{z}$$ which establishes a one to one correspondence between the nonzero points of the $z$ and $w$ planes. Since $z\overline{z} = |z|^2$, the mapping can be described by means of the successive transformations $$g(z)=\frac{z}{|z|^2},\quad f(z)=\overline{g(z)}.$$ The first transformation $g(z)$ is an inversion with respect to the unit circle $|z| = 1$. That is, the image of a nonzero point $z$ is the point $g(z)$ with the properties $$|g(z)| = \frac{1}{|z|}\quad\text{and}\quad \textbf{arg } g(z) = \textbf{arg } z.$$ Thus the points exterior to the circle $|z| = 1$ are mapped onto the nonzero points interior to it, and conversely. Any point on the circle is mapped onto itself. The second transformation $f(z)=\overline{g(z)}$ is simply a reflection in the real axis.

If we consider the function \begin{eqnarray*} T(z)=\frac{1}{z}, \quad z\neq 0, \end{eqnarray*} we can define $T$ at the origin and at the point at infinity so as to be continuous on the extended complex plane. In order to make $T$ continuous on the extended plane, then, we write \begin{eqnarray*} T(0)=\infty,\quad T(\infty)=0, \quad \text{and}\quad T(z)=\frac{1}{z} \end{eqnarray*} for the remaining values of $z$.

## Mappings by $1/z$

An interesting property of the mapping $w = 1/z$ is that it transforms circles and lines into circles and lines.

You can observe this intuitively in the following applet. Choose a Line or Circle. Drag points around on the left-side, you can also drag the line or circle. Observe what happens to the image of each point on the right-side window.

Sorry, the applet is not supported for small screens. Rotate your device to landscape. Or resize your window so it's more wide than tall.

When $A$, $B$, $C$ and $D$ are all real numbers satisfying the condition $B^2+C^2>4AD$, the equation \begin{eqnarray}\label{circle01} A\left(x^2+y^2\right)+Bx+Cy+D=0 \end{eqnarray} represents and arbitrary circle or line, where $A\neq 0$ for a circle and $A=0$ for a line.

By using the method of completing the squares, we can rewrite equation (\ref{circle01}) as follows \begin{eqnarray*}\label{circle02} \left(x+\frac{B}{2A}\right)+\left(y+\frac{C}{2A}\right)=\left(\frac{\sqrt{B^2+C^2-4AD}}{2A}\right)^2 \end{eqnarray*} This makes evident the need for condition $B^2+C^2>4AD$ when $A\neq 0$. When $A = 0$, the condition becomes $B^2 + C^2 > 0$, which means that $B$ and $C$ are not both zero.

Now, using the relations \begin{eqnarray*} x=\frac{z+\overline{z}}{2},\quad y=\frac{z-\overline{z}}{2i}, \end{eqnarray*} we can rewrite equation (\ref{circle01}) in the form \begin{eqnarray}\label{circle03} 2Az\overline{z}+(B-iC)z+(B+iC)\overline{z}+2D=0. \end{eqnarray} Since $w=1/z$, equation (\ref{circle03}) becomes \begin{eqnarray*} 2Dw\overline{w}+(B+iC)w+(B-iC)\overline{w}+2A=0 \end{eqnarray*} and using the relations \begin{eqnarray*} u=\frac{w+\overline{w}}{2},\quad v=\frac{w-\overline{w}}{2i}, \end{eqnarray*} we obtain \begin{eqnarray}\label{circle04} D(u^2+v^2)+Bu-Cv+A=0 \end{eqnarray} which also represents a circle or line.

It is now clear from equations (\ref{circle01}) and (\ref{circle04}) that

1. a circle ($A \neq 0$) not passing through the origin ($D \neq 0$) in the $z$ plane is transformed into a circle not passing through the origin in the $w$ plane;
2. a circle ($A \neq 0$) through the origin ($D = 0$) in the $z$ plane is transformed into a line that does not pass through the origin in the $w$ plane;
3. a line ($A = 0$) not passing through the origin ($D \neq 0$) in the $z$ plane is transformed into a circle through the origin in the w plane;
4. a line ($A = 0$) through the origin $(D = 0)$ in the $z$ plane is transformed into a line through the origin in the $w$ plane.

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