What does $\left| z - 3 \right| = 1$ look like ? A circle of radius 1, centered at 3.

What does $\displaystyle \left| \frac{1}{z} - 3 \right| = 1$ look like ?

$$\begin{aligned} \left| \frac{1}{z} - 3 \right| &= 1 \\ \left| \frac{1-3z}{z} \right|^2 &= 1 \\ \left| 1-3z \right|^2 &= \left| z \right|^2 \\ (1-3z)\overline{(1-3z)} &= z \overline z \\ (1-3z)(1-3\overline z) &= z \overline z \\ 1-3z-3\overline z + 9z\overline z &= z \overline z \\ 8z\overline z - 3z - 3\overline z + 1 &= 0 \\ z\overline z - \frac{3}{8}z - \frac{3}{8} \overline z + \frac{1}{8} &= 0 \\ (z - \frac{3}{8})(\overline z - \frac{3}{8}) - \frac{3^2}{8^2} + \frac{1}{8} &= 0 \\ \left| z - \frac{3}{8} \right|^2 &= \frac{1}{64} \\ \left| z - \frac{3}{8} \right| &= \frac{1}{8} \\ \end{aligned}$$

$\therefore$ A circle of radius $\displaystyle \frac{1}{8}$, centered at $\displaystyle \frac{3}{8}$.

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 3:26 AM