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.

Question

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?

Answer 1

The problem is slightly misstated since the word "variable" is missing. It should be like this

If $a$ is a fixed complex number and $b$ is a variable real number, then prove that the equation $a\bar z+\bar a z+b=0$ is a straight line passing though origin.

Write $\,a = u+vi\,$ and $\,z = x+yi.\,$ Then $$ a\bar z+\bar a z = (ux+vy) + (vx-uy)i $$ and if this number is equal to a real number $\,b\,$ then $\,vx-uy=0,\,$ which is the equation of a line through the origin.

Answer 2

$$ \begin{aligned} a\bar{z}&=|a||z|\angle \left(arg(a)-arg(z)\right)\\ \bar{a}z&=|a||z|\angle \left(arg(z)-arg(a)\right)\\ \\ a\bar{z}+\bar{a}z+b&=2|a||z|\cos{\left(arg(a)-arg(z)\right)}+b \end{aligned} $$

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 $b=0$