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…
Best Answer
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 $x=0$