If $\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi }{2}$where z is a complex number, locus of z is 

Let $z=x+iy$

$\begin{array}{l}\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi }{2}\Rightarrow \arg \left(\frac{(x-1)+iy}{(x+1)+iy}\right)=\frac{\pi }{2}\\ \therefore \arg \left[\frac{\left(\right(x-1)+iy)(x+1-iy)}{(x+1{)}^2+y^2}\right]=\frac{\pi }{2}\\ \Rightarrow \arg \left[x^2-1+y^2+i(xy+y-xy+y)\right]=\frac{\pi }{2}\\ \Rightarrow \arg \left[x^2-1+y^2+2iy\right]=\frac{\pi }{2}\\ \Rightarrow x^2+y^2-1=0\end{array}$

$\Rightarrow$ which is the unit circle. $\left|Z\right|$= 1 

But drawing the circle we note that points lying on the portion of the circle above the x-axis only 

satisfy the condition $\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi }{2}$, where, as

the points on this circle lying below the x-axis

satisfy the condition $\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi }{2}$, Hence

the locus is the portion of the circle $\left|Z\right|$ = 1 lying above the x-axis excluding z = ±1