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?
Suppose that, in this question, the principal value of the argument is chosen to be the unique value of the argument that lies within the interval $(−\pi,\pi]$. Then, since, for any $z$ for which $\arg(z\pm i)$ are defined, $\arg(z-i)-\arg(z+i)\lt\frac{\pi}{2}$ holds, there is no $z$ such that $\arg(z-i)-\arg(z+i)=\frac{2\pi}{3}$ (for the details, see the end of this answer).
This suggests that the question defines the range of the principal value as being in the closed-open interval $[0, 2\pi)$.
Let $z=x+yi$ where $x,y\in\mathbb R$.
Then, we have $$\arg(x+yi)=\begin{cases}2\pi-\arctan(\frac{-y}{x})&\text{if $x\gt 0$ and $y\lt 0$} \\\arctan(\frac yx)&\text{if $x\gt 0$ and $y\geqslant 0$} \\\frac{3}{2}\pi&\text{if $x=0$ and $y\lt 0$} \\\text{undefined}&\text{if $x=0$ and $y=0$} \\\frac{\pi}{2}&\text{if $x=0$ and $y\gt 0$} \\\pi+\arctan(\frac{-y}{-x})&\text{if $x\lt 0$ and $y\leqslant 0$} \\\pi-\arctan(\frac{y}{-x})&\text{if $x\lt 0$ and $y\gt 0$} \end{cases}$$ i.e. $$\arg(x+yi)=\begin{cases}2\pi+\arctan(\frac{y}{x})&\text{if $x\gt 0$ and $y\lt 0$} \\\arctan(\frac yx)&\text{if $x\gt 0$ and $y\geqslant 0$} \\\frac{3}{2}\pi&\text{if $x=0$ and $y\lt 0$} \\\text{undefined}&\text{if $x=0$ and $y=0$} \\\frac{\pi}{2}&\text{if $x=0$ and $y\gt 0$} \\\pi+\arctan(\frac yx)&\text{if $x\lt 0$}\end{cases}$$
Since we have $$\arg(x+(y-1)i)=\begin{cases} 2\pi+\arctan(\frac{y-1}{x})&\text{if $x\gt 0$ and $y\lt 1$} \\\arctan(\frac{y-1}x)&\text{if $x\gt 0$ and $y\geqslant 1$} \\\frac{3}{2}\pi&\text{if $x=0$ and $y\lt 1$} \\\text{undefined}&\text{if $x=0$ and $y=1$} \\\frac{\pi}{2}&\text{if $x=0$ and $y\gt 1$} \\\pi+\arctan(\frac{y-1}x)&\text{if $x\lt 0$} \end{cases}$$ and $$\arg(x+(y+1)i)=\begin{cases} 2\pi+\arctan(\frac{y+1}{x})&\text{if $x\gt 0$ and $y\lt -1$} \\\arctan(\frac{y+1}x)&\text{if $x\gt 0$ and $y\geqslant -1$} \\\frac{3}{2}\pi&\text{if $x=0$ and $y\lt -1$} \\\text{undefined}&\text{if $x=0$ and $y=-1$} \\\frac{\pi}{2}&\text{if $x=0$ and $y\gt -1$} \\\pi+\arctan(\frac{y+1}x)&\text{if $x\lt 0$} \end{cases}$$ we get $$\arg(x+(y-1)i)-\arg(x+(y+1)i)$$ $$=\begin{cases} 2\pi+\arctan(\frac{y-1}{x})-(2\pi+\arctan(\frac{y+1}{x}))&\text{if $x\gt 0$ and $y\lt -1$} \\2\pi+\arctan(\frac{y-1}{x})-\arctan(\frac{y+1}x)&\text{if $x\gt 0$ and $-1\leqslant y\lt 1$} \\\arctan(\frac{y-1}x)-\arctan(\frac{y+1}x)&\text{if $x\gt 0$ and $y\geqslant 1$} \\\frac{3}{2}\pi-\frac{3}{2}\pi&\text{if $x=0$ and $y\lt -1$} \\\text{undefined}&\text{if $x=0$ and $y=-1$} \\\frac{3}{2}\pi-\frac{\pi}{2}&\text{if $x=0$ and $-1\lt y\lt 1$} \\\text{undefined}&\text{if $x=0$ and $y=1$} \\\frac{\pi}{2}-\frac{\pi}{2}&\text{if $x=0$ and $y\gt 1$} \\\pi+\arctan(\frac{y-1}x)-(\pi+\arctan(\frac{y+1}x))&\text{if $x\lt 0$} \end{cases}$$
$$=\begin{cases} \arctan(\frac{y-1}{x})-\arctan(\frac{y+1}{x})\lt 0&\text{if $x\gt 0$ and $y\lt -1$} \\2\pi+\arctan(\frac{y-1}{x})-\arctan(\frac{y+1}x)\gt 2\pi&\text{if $x\gt 0$ and $-1\leqslant y\lt 1$} \\\arctan(\frac{y-1}x)-\arctan(\frac{y+1}x)\lt 0&\text{if $x\gt 0$ and $y\geqslant 1$} \\0&\text{if $x=0$ and $y\lt -1$} \\\text{undefined}&\text{if $x=0$ and $y=-1$} \\\pi&\text{if $x=0$ and $-1\lt y\lt 1$} \\\text{undefined}&\text{if $x=0$ and $y=1$} \\ 0&\text{if $x=0$ and $y\gt 1$} \\\arctan(\frac{y-1}x)-\arctan(\frac{y+1}x)&\text{if $x\lt 0$} \end{cases}$$
So, under the condition $x\lt 0$, we have $$\arctan\bigg(\frac{y-1}x\bigg)-\arctan\bigg(\frac{y+1}x\bigg)= \frac{2\pi}{3}$$ i.e. $$\arctan\left(\frac{2x}{1-x^2-y^2}\right)=-\frac{\pi}{3},$$ i.e. $$\frac{2x}{1-x^2-y^2}=-\sqrt 3,$$ i.e. $$\left(x-\frac{1}{\sqrt 3}\right)^2+y^2=\left(\frac{2}{\sqrt 3}\right)^2\tag1$$
Since the region we seek is the inside of the circle $(1)$ with $x\lt 0$, the area is given by
$$\pi\bigg(\frac{2}{\sqrt 3}\bigg)^2\times\frac{\frac{2}{3}\pi}{2\pi}-\frac 12\bigg(\frac{2}{\sqrt 3}\bigg)^2\sin\bigg(\frac{2\pi}{3}\bigg)=\color{red}{\frac{4}{9}\pi-\frac{1}{\sqrt 3}}$$
In the following, let us prove that if the principal value of the argument is chosen to be the unique value of the argument that lies within the interval $(−\pi,\pi]$, then, for any $z$ for which $\arg(z\pm i)$ are defined, $\arg(z-i)-\arg(z+i)\lt\frac{\pi}{2}$ holds.
Proof :
Let $z=x+yi$ where $x,y\in\mathbb R$.
If the principal value of the argument is chosen to be the unique value of the argument that lies within the interval $(−\pi,\pi]$, then we have $$\arg(x+yi)=\begin{cases}\text{arctan}(\frac yx)&\text{if $x\gt 0$ and $y\geqslant 0$} \\-\arctan(\frac{-y}{x})&\text{if $x\gt 0$ and $y\lt 0$} \\\frac{\pi}{2}&\text{if $x=0$ and $y\gt 0$} \\\text{undefined}&\text{if $x=0$ and $y=0$} \\-\frac{\pi}{2}&\text{if $x=0$ and $y\lt 0$} \\\pi-\arctan(\frac{y}{-x})&\text{if $x\lt 0$ and $y\geqslant 0$} \\-(\pi-\arctan(\frac{-y}{-x}))&\text{if $x\lt 0$ and $y\lt 0$}\end{cases}$$ i.e. $$\arg(x+yi)=\begin{cases}\text{arctan}(\frac yx)&\text{if $x\gt 0$} \\\frac{\pi}{2}&\text{if $x=0$ and $y\gt 0$} \\\text{undefined}&\text{if $x=0$ and $y=0$} \\-\frac{\pi}{2}&\text{if $x=0$ and $y\lt 0$} \\\arctan(\frac{y}{x})+\pi&\text{if $x\lt 0$ and $y\geqslant 0$} \\\arctan(\frac yx)-\pi&\text{if $x\lt 0$ and $y\lt 0$} \end{cases}$$
Since we have $$\arg(x+(y-1)i)=\begin{cases}\text{arctan}(\frac{y-1}x)&\text{if $x\gt 0$} \\\frac{\pi}{2}&\text{if $x=0$ and $1\lt y$} \\\text{undefined}&\text{if $x=0$ and $y=1$} \\-\frac{\pi}{2}&\text{if $x=0$ and $y\lt 1$} \\\arctan(\frac{y-1}{x})+\pi&\text{if $x\lt 0$ and $1\leqslant y$} \\\arctan(\frac{y-1}x)-\pi&\text{if $x\lt 0$ and $y\lt 1$} \end{cases}$$
and
$$\arg(x+(y+1)i)=\begin{cases}\text{arctan}(\frac{y+1}x)&\text{if $x\gt 0$} \\\frac{\pi}{2}&\text{if $x=0$ and $-1\lt y$} \\\text{undefined}&\text{if $x=0$ and $y=-1$} \\-\frac{\pi}{2}&\text{if $x=0$ and $y\lt -1$} \\\arctan(\frac{y+1}{x})+\pi&\text{if $x\lt 0$ and $-1\leqslant y$} \\\arctan(\frac{y+1}x)-\pi&\text{if $x\lt 0$ and $y\lt -1$} \end{cases}$$
we get $$\arg(x+(y-1)i)-\arg(x+(y+1)i)$$ $$=\begin{cases}\text{arctan}(\frac{y-1}x)-\text{arctan}(\frac{y+1}x)&\text{if $x\gt 0$} \\-\frac{\pi}{2}-(-\frac{\pi}{2})&\text{if $x=0$ and $y\lt -1$} \\\text{undefined}&\text{if $x=0$ and $y=-1$} \\-\frac{\pi}{2}-\frac{\pi}{2}&\text{if $x=0$ and $-1\lt y\lt 1$} \\\text{undefined}&\text{if $x=0$ and $y=1$} \\\frac{\pi}{2}-\frac{\pi}{2}&\text{if $x=0$ and $1\lt y$} \\\arctan(\frac{y-1}x)-\pi-(\arctan(\frac{y+1}x)-\pi)&\text{if $x\lt 0$ and $y\lt -1$} \\\arctan(\frac{y-1}x)-\pi-(\arctan(\frac{y+1}{x})+\pi)&\text{if $x\lt 0$ and $-1\leqslant y\lt 1$} \\\arctan(\frac{y-1}{x})+\pi-(\arctan(\frac{y+1}{x})+\pi)&\text{if $x\lt 0$ and $1\leqslant y$}\end{cases}$$
$$=\begin{cases}\text{arctan}(\frac{y-1}x)-\text{arctan}(\frac{y+1}x)\lt 0&\text{if $x\gt 0$} \\0&\text{if $x=0$ and $y\lt -1$} \\\text{undefined}&\text{if $x=0$ and $y=-1$} \\-\pi&\text{if $x=0$ and $-1\lt y\lt 1$} \\\text{undefined}&\text{if $x=0$ and $y=1$} \\0&\text{if $x=0$ and $1\lt y$} \\\arctan(\frac{y-1}x)-\arctan(\frac{y+1}x)\lt\frac{\pi}{2}&\text{if $x\lt 0$ and $y\lt -1$} \\-2\pi+\arctan(\frac{y-1}x)-\arctan(\frac{y+1}{x})\lt 0&\text{if $x\lt 0$ and $-1\leqslant y\lt 1$} \\\arctan(\frac{y-1}{x})-\arctan(\frac{y+1}{x})\lt\frac{\pi}{2}&\text{if $x\lt 0$ and $1\leqslant y$}\end{cases}$$
Therefore, we can say that, for any $z$ for which $\arg(z\pm i)$ are defined, $\arg(z-i)-\arg(z+i)\lt\frac{\pi}{2}$ holds.$\quad\blacksquare$
