If $a$ is a complex number and b is a real number, then prove that the equation $a\bar z+\bar a z+b=0$ is a straight line passing though origin. Generally, $z\bar z +\bar a z + a\bar z+b=0$ represents a circle 
The lack of the term $z\bar z$ confirms that it is a straight line. 
But we still have the term b, which is a real number and isn’t necessarily zero, then how do we prove that it passes though the origin?

$$
\begin{aligned}
a\bar{z}&=|a||z|\angle \left(arg(a)-arg(z)\right)\\
\bar{a}z&=|a||z|\angle \left(arg(z)-arg(a)\right)\\
\\
a\bar{z}+\bar{a}z+b&=2|a||z|\cos{\left(arg(a)-arg(z)\right)}+b
\end{aligned}
$$
Some interpretations: 
$z$ is a set of points in the complex plane such that $\vec{Oz}\cdot\vec{Oa}=\frac{b}{2}$.
$z$ lies on a line that is perpendicular to $\vec{Oa}$
$z=\frac{ab}{2|a|^{2}}+i(ma), m\in\mathbb{Z}$
The line only goes through origin if $b=0$
