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?

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.
