How to find the locus of points in the argand plane represented as follows: A set of points in the argand plane is given by the equation:
$$arg\left( \frac{z-6-3i}{z-3-6i}\right)={\pi \over4}$$
How do I find the equivalent locus in the cartesian x-y plane. The standard rule is to put $z=x+iy$ and solve the given equation for a relation in $x$ and $y$. However the above equation leads to two linear equations in $x$ and $y$. What do I interpret from these lines.
Intuitively, it appears that the locus should be a closed figure such that the line joining the points represented by $3+6i$ and $6+3i$ subtends an angle of ${\pi \over4}$ on each point of it.. However, this does not appear on solving the equation. I solved it using the fact that the directions of the position vectors represented by the points in numerator and denominator are inclined at ${\pi \over4}$

Let's make a consideration first of all:
$$\forall\ x+iy\in\mathbb{C}-\{0+0i\} $$
$$Arg(x+iy)=\pi/4\ \Leftrightarrow 
\begin{cases}
  y>0\\
  x=y\\
\end{cases} \Leftrightarrow 
\begin{cases}
  x>0\\
  x=y\\
\end{cases}$$
The proof of these property is direct if you think about the sinus and the cosine of 45°.
We are going to use use the second one,but the result should be the same with the third one.
Well let's start:
$$ \frac{z-6+3i}{z-3-6i}= 1\ +\ \frac{-3+3i}{z-3-6i}$$
$$ let's\ suppose\ z=x+iy$$
$$1\ +\ \frac{-3+3i}{x + iy-3-6i}= 1\ +\frac{-3+3i}{x + iy-3-6i} \left(\frac{x-iy-3+6i}{x-iy-3+6i}\right) = $$
$$= 1 + \frac{-3x+3y-9}{(x-3)^2 + (y-6)^2}\ +\ i\left(\frac{3x+3y-27}{(x-3)^2 + (y-6)^2}\right)$$
If I've done the calculus right.Then let's use the proposition showed before:
$$
\begin{cases}
  \frac{3x+3y-27}{(x-3)^2 + (y-6)^2}>0\\
  1 + \frac{-3x+3y-9}{(x-3)^2 + (y-6)^2}=\frac{3x+3y-27}{(x-3)^2 + (y-6)^2}\\
\end{cases}
$$
Then doing the calculus should becomes:
$$
\begin{cases}
  y> x+3\\
  x^2\ +y^2\ -12x\ -12y\ +\ 63\ =\ 0\\
\end{cases} 
$$
But we have that:
$$ x^2\ +y^2\ -12x\ -12y\ +\ 63\ =\ 0\ \Leftrightarrow\ (x-6)^2\ +\ (y-6)^2\ =9$$
Which is a circle,then the solution is the arc of circle inside the half plan y>x+3.

The blue circle is the result of the third method.
