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202212281117 Problem 1.18

Show that an equation for the circle C(z_0,r) is

z\overline{z}-\overline{z}_0z-z_0\overline{z}+z_0\overline{z}_0=r^2.

Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.

Roughwork.

Writing

\begin{aligned} |z-z_0| & = r \\ |(a+bi)-(a_0+b_0i)| & = r \\ |(a-a_0)+(b-b_0)i| & = r \\ (a-a_0)^2 + (b-b_0)^2 & = r^2 \\ \end{aligned}

and

\begin{aligned} &\quad\enspace z\overline{z}-\overline{z}_0z-z_0\overline{z}+z_0\overline{z}_0 \\ & = (a+bi)(a-bi) - (a_0-b_0i)(a+bi) \\ & \qquad\qquad -(a_0+b_0i)(a-bi)+(a_0+b_0i)(a_0-b_0i) \\ & = \cdots \\ \end{aligned}

is to let ends meet, without use of properties of conjugates:

\begin{aligned} |\overline{z}| & = |z| \\ \mathrm{arg}\overline{z} & = -\mathrm{arg}z \\ \overline{z_1+z_2} & = \overline{z}_1+\overline{z}_2 \\ \overline{z_1-z_2} & = \overline{z}_1-\overline{z}_2 \\ \overline{z_1z_2} & = \overline{z}_1\overline{z}_2 \\ \mathrm{Re}z & = (z+\overline{z})/2 \\ \mathrm{Im}z & = (z-\overline{z})/2i \\ z\overline{z} & = |z|^2 \\ \end{aligned}

(to be continued)