complex numbers
Find the area of region enclosed by the locus of $z$ given by $\arg(z-i) – \arg(z+i)= \frac{2\pi}{3}$ and imaginary axis (where $i= \sqrt {-1}$)
What I did was I put $$\tan (\alpha) =z-i$$ and $$\tan (\beta) = z+i$$ and solving for $\tan(\alpha-\beta) $ I got the locus of $z$ as $xy=\frac {1}{\sqrt 3}$ but now how to find area?
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 $x=0$
As seen from the diagram, the inner angle $\alpha$ of the triangle is the difference between the outer angle $\arg(z-1)$ and the other inner angle $\arg(z)$, i.e.
$$\alpha = \arg(z-1) - \arg(z) = \arg\frac{z-1}z$$
For constant $\alpha$, the vertex of $\alpha$ follows the circumference of the red circle, but only above the $x$-axis. It can be shown similarly that the subtended angle below the $x$-axis is $\arg\frac{z}{z-1}\ne \alpha$, thus excluded.
Best Answer
Let $z=x+yi$ where $x,y\in\mathbb R$.
We have, using this, $$\begin{align}&\text{Arg}(x+(y-1)i)-\text{Arg}(x+(y+1)i)\\\\&=\begin{cases}\arctan\left(\frac{y-1}{x}\right)-\arctan\left(\frac{y+1}{x}\right)&\text{if$\ x\gt 0\ \text{or}\ (x\lt 0\ \text{and}\ y\ge 1)\ \text{or}\ (x\lt 0\ \text{and}\ y\lt -1)$}\\\\\arctan\left(\frac{y-1}{x}\right)-\arctan\left(\frac{y+1}{x}\right)-2\pi&\text{if$\ x\lt 0\ \text{and}\ -1\le y\lt 1$}\\\\\end{cases}\end{align}$$
Since the latter case cannot be equal to $\frac 23\pi$, we have
$$\begin{align}\frac{2\pi}{3}&=\text{Arg}(x+(y-1)i)-\text{Arg}(x+(y+1)i)\\\\&=\arctan\left(\frac{y-1}x\right)+\arctan\left(\frac{-y-1}x\right)\\\\&=\arctan\left(\frac{2x}{1-x^2-y^2}\right)+\pi\end{align}$$ where $$x\gt 0\quad \text{or}\quad\ (x\lt 0\ \text{and}\ y\ge 1)\quad \text{or}\quad (x\lt 0\ \text{and}\ y\lt -1)$$ So, we have$$\arctan\left(\frac{2x}{1-x^2-y^2}\right)=-\frac{\pi}{3},$$ i.e. $$\frac{2x}{1-x^2-y^2}=-\sqrt 3,$$ i.e. $$\left(x-\frac{1}{\sqrt 3}\right)^2+y^2=\left(\frac{2}{\sqrt 3}\right)^2\tag1$$
Therefore, the region we seek is the inside of the circle $(1)$ with $x\ge 0$.
So, the area is $$\pi\left(\frac{2}{\sqrt 3}\right)^2\cdot \frac{2\pi-2\cdot\frac{\pi}{3}}{2\pi}+2\times\frac 12\times\frac{1}{\sqrt 3}\times 1=\color{red}{\frac{8}{9}\pi+\frac{1}{\sqrt 3}}$$