Finding the locus of a complex number Find the locus of $\arg\left(\frac{z-3}{z}\right) = \frac{\pi}{4}$ where $z$ represent complex number.
Working: $\arg\left(\frac{z-3}{z}\right) $ can be written as $\arg(z-3)-\arg(z) = \frac{\pi}{4}$, or $\arg\left((x-3)+iy\right) - \arg(x+iy)=\frac{\pi}{4}$.
If we take tangent to both side we get :
$$
\begin{align*}
\tan \left[\arg\left((x-3)+iy\right) -\arg (x+iy)\right] = \tan \frac{\pi}{4},\\
\tan\left[\frac{\arg\left((x-3)+iy\right) - \arg(x+iy)}{1+ \arg\left((x-3)+iy\right) \arg(x+iy)}\right] = 1.
\end{align*}
$$
Please suggest further...
 A: So, $$\arctan \left(\frac y{x-3}\right)-\arctan \left( \frac yx\right)=\frac\pi4$$
$$\implies \arctan\left(\frac{ \frac y{(x-3)}-\frac yx}{1+ \frac y{(x-3)}\frac yx}\right)=\frac\pi4$$ 
$$\implies \frac{3y}{(x-3)x+y^2}=\tan\left(\frac\pi4\right)=1$$ assuming $x(x-3)\ne0$ 
$$\implies x^2-3x+y^2=3y$$
If $x=3,\arctan \left(\frac y{x-3}\right)-\arctan \left( \frac yx\right)=\frac\pi4$ reduces to $$\arctan \left( \frac y0\right)-\arctan \left( \frac y3\right)=\frac\pi4$$
$$\text{Now, }\arctan \left( \frac y0\right)= \begin{cases}
   \ \frac{\pi}2 & \text{if } y>0 \\
    \ -\frac{\pi}2 & \text{if } y<0\\
     \text{ undefined}       & \text{if } y= 0
  \end{cases}$$
$$\text{ If }y>0, \arctan \left( \frac y3\right)=\frac\pi2-\frac\pi4=\frac\pi4$$
$$\implies \frac y3=\tan\left(\frac\pi4\right)=1\implies y=3\text{ at }x=3$$
$$\text{ If }y<0, \arctan \left( \frac y3\right)=-\frac\pi2-\frac\pi4=-\frac{3\pi}4$$
$$\implies \frac y3=\tan\left(-\frac{3\pi}4\right)=\tan\left(\pi-\frac{\pi}4\right)=-\tan\left(\frac\pi4\right)=-1\implies y=-3\text{ at }x=3$$
Similarly, for $x=0$
A: $\arg(\dfrac{z-3}{z})=\dfrac{\pi}{4}\implies\Re(\dfrac{z-3}{z})=\Im(\dfrac{z-3}{z})$ and$\Re(\dfrac{z-3}{z})>0,z\neq0 $$\implies\Re(\dfrac{|z|^2-3\bar z}{|z|^2} )=\Im(\dfrac{|z|^2-3\bar z}{|z|^2})$$\implies x^2+y^2-3x=3y,y>0$(As$\Im(\dfrac{z-3}{z})=\dfrac{3y}{x^2+y^2}>0 $ if and only if $y>0$)
