Here is a complex analysis homework problem I can't quite figure out:
Prove that every line or circle in $\mathbb{C}$ is the solution set of an equation of the form $a|z|^2+\bar{w}z+w\bar{z}+b=0$, where $a,b\in\mathbb{R}$ and $w,z\in\mathbb{C}$. Conversely, show that every equation of this form has a line, circle, point, or the empty set as its solution set.
So far, I've tried to rewrite the equation of a line in $\mathbb{R}^2$ as $y=mx+b$ in $\mathbb{C}$, where $m$ is real and $x,b$ are complex. I know that a circle in the complex plane is given by $|z-a|=r$, where $a$ is the center and $r$ is the radius.
I also noticed that $\bar{w}z+w\bar{z}=2\text{Re}(\bar{w}z)$. I'm just not sure how all these pieces fit together in answering the question. Any help would be greatly appreciated.
Best Answer
You need to use the property of complex conjugation to express the circle equation
\begin{align} |z - z_0|^2 &= r^2 \Leftrightarrow \overline{(z - z_0)}(z-z_0) = r^2 \Leftrightarrow (\overline{z} - \overline{z_0})(z - z_0) = r^2 \Leftrightarrow \\ & z\overline{z} - \overline{z}z_0 - z\overline{z_0} + |z_0|^2 = r^2 \end{align}
Now make the substitution $z_0 = -\frac{\overline{\alpha}}{A}$ where $A \in \mathbb{R} \setminus \{0\}$
$$ z\overline{z} + \overline{z}\frac{\overline{\alpha}}{A} + z\frac{\alpha}{A} + \left|\frac{\overline{\alpha}}{A}\right|^2 - r^2 = 0 $$
Mutliplying the equation by $A$
$$ Az\overline{z} + \overline{z\alpha} + z\alpha + A\left(\left|\frac{\alpha}{A}\right|^2 - r^2\right) = 0 $$
And setting $B = A\left(\left|\frac{\alpha}{A}\right|^2 - r^2\right)$ yields the general form of the equation for a circle in the complex plane. This equation also describes lines which can be viewed as circles with infinite radius.
$$ Az\overline{z} + \overline{z\alpha} + z\alpha + B = 0 $$
When $A = 0$ it represents a line,
$$ \overline{z\alpha} + z\alpha + B = 0 $$
You can convice yourself that this equation describes a line by setting $z = x + iy$ and $\alpha = p + iq$.