<s>[INST] 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...
 [/INST] $\arg(\dfrac{z-3}{z})=\dfrac{\pi}{4}\implies\Re(\dfrac{z-3}{z})=\Im(\dfrac{z-3}{z})$ 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 $y>0$)
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