Area of region enclosed by the locus of a complex number 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?

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 
$$\pi\left(\frac{2}{\sqrt 3}\right)^2\cdot \frac{2\pi-2\cdot\frac{\pi}{3}}{2\pi}+2\times\frac 12\times\frac{1}{\sqrt 3}\times 1=\color{red}{\frac{8}{9}\pi+\frac{1}{\sqrt 3}}$$
