Show that the equation $b(z+\bar z)+\bar b(\bar z-z)+a=0$ represents two straight lines in the complex plane

Question

Q:if $a$ is real, $b=p+iq,z=x+iy$ and $\bar b=p-iq,\bar z=x-iy$ then show that the equation $b(z+\bar z)+\bar b(\bar z-z)+a=0$ represents two straight lines in the complex plane.
My approach:Simplify the equation i get,\begin{align}b(z+\bar z)+\bar b(\bar z-z)+a & =0 \\ (p+iq)(x+iy+x-iy)+(p-iq)(x-iy-x-iy)+a & =0 \\ (2xp-2yq+a)+i(2xq-2py) & =0 \end{align} But now i get stuck because i don't know how to show that it represent two straight lines. Any hints or solution will be appreciated.
Thanks in advance.

Answer 1

You have yielded :

$$(2xp-2yq+a)+i(2xq-2py) =0$$

Now, for a complex expression $x_1 + y_1 i$ to be equal to zero, both imaginary and real parts must be equal to zero. Thus :

$$2xp-2yq+a = 0 \implies y = \frac{p}{q}x + \frac{a}{2q}$$

$$2xq - 2py = 0 \implies y = \frac{q}{p}x$$

which are both equations of straight lines, as $p,q,a$ are constants $ \in \mathbb R$.

Answer 2

Hint:

1. If $A$ and $B$ are real, $A+Bi=0$ if and only if $A=B=0$.
2. $2xp-2yq+a$ and $2xq-2py$ are real.