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 ${\pi \over4}$
 A: Instead of substituting $z=x+iy$ and making it all messy, here is a simple geometric solution:
As you said, the vectors represented by the points in the numerator and denominator contain the constant angle $\pi\over4$.
This means that the locus should be an arc of a circle in which the chord made by joining the 2 given points subtends an angle of $\pi\over4$ on the arc.

Since the chord subtends 90 degrees at the centre, the given points make it very easy to find the centre because it must be directly above (6,3) and directly rightward of (3,6). So it is at (6,6). Then the radius is obviously 3. The cartesian equation for such circle is $$(x-6)^2+(y-6)^2=9$$
But since we want only the part on the right of the chord, we add the additional constraint $$x+y>9$$ (because $x+y=9$ is the equation of the chord.)
These two relations in $x$ and $y$ simultaneously can represent the equivalent of the complex locus given in the Cartesian plane. 
A: 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.
