Analytical Geometry problem with complex numbers - alternate solutions. The question is to show that the equation of the lines making angles $45^\circ$ with the line:
$$ \bar{a}z + a\bar{z} + b = 0; \;\;\;\;\; a,z \in \mathbb{C}, b \in \mathbb{R} $$
and passing through a point $c \in \mathbb{C}$ is:
$$ \dfrac{z-c}{a} \pm i\dfrac{\bar{z} - \bar{c} } { \bar{a} } = 0 $$
Now, I know one method to solve this problem. Taking $z = x + iy$ and finding the slope of the given line, and then finding the slopes of the required lines...
But that is way too long! Are there better, more elegant methods?
 A: Drawing figures will help you understand my answer.
Let $L$ be the given line. Let $D(d)$ be the intersection point of $L$ and the perpendicular line of $L$. Noting that the normal vector to $L$ is $a$, there exists $k\in\mathbb R$ such that
$$d+ka=c\iff d=c-ka.$$
In the following, we suppose that $k\not=0.$ The case $k=0$ means that $c$ is on the $L$. This case is easy so I'll mention about this case at the last.
Now, let $L_1,L_2$ be the lines we want, and let $E(e), F(f)$ be the intersection point of "$L$ and $L_1$" and "$L$ and $L_2$" respectively.
Hence, we have
$$e=d+iak, f=d-ika.$$
(Do you see why? Again, drawing figures will help you.)
Noting that the line which passes through $\alpha,\beta$ represents
$$(\bar{\beta}-\bar{\alpha})z-(\beta-\alpha)\bar{z}=\bar{\beta}\alpha-\beta\bar{\alpha},$$
we know $L_1$ is represented as
$$(\bar{e}-\bar{c})z-(e-c)\bar{z}=\bar ec-e\bar c$$
$$\iff (\bar d-\bar aki-\bar{c})z-(d+iak-c)\bar{z}=(\bar d-\bar aki)c-(d+iak)\bar c$$
Setting $\bar d=\bar c-k\bar a$ in this and dividing the both sides by $k(\not=0)$ gives us 
$$\bar a(1+i)(z-c)+a(i-1)(\bar z-\bar c)=0$$
Dividing the both sides by $a\bar a(1+i)$ gives us
$$\frac{z-c}{a}+\frac{\bar z-\bar c}{\bar a}\times\frac{i-1}{1+i}=0.$$
Noting $(i-1)/(1+i)=i$, this is what we want.
We can repeat the same argument as above for $L_2$, so we can prove it for the $k\not=0$ case.
Finally, I'm going to mention the $k=0$ case. This means that $c$ is on $L$. Hence, $L_1$ is a line which passes through $c$ and $c+a-ia.$ So, we can repeat the same argument as above. Q.E.D.
