$P$ is a point denoting $z$ in the argand diagram and if $\frac{z-i}{z-1}$ is always purely imaginary, then…
$P$ is a point denoting $z$ in the argand diagram and if $\frac{z-i}{z-1}$ is always purely imaginary, then locus of $P$ is
the circle with centre $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$
the circle with centre $\left(-\frac{1}{2},-\frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$
the points on the circle with centre $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$, excluding the points $(1,0)$ and $(0,1)$
the points on the circle with centre $\left(-\frac{1}{2},-\frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$, excluding the origin
Solution
Let $z=x+i y$, then $\frac{z-i}{z-1}=\frac{x+i(y-1)}{(x-1)+i y}$
So, $\quad \operatorname{Re}\left(\frac{z-i}{z+1}\right)=\frac{x(x-1)+y(y-1)}{(x-1)^2+y^2}$
$\because \frac{z-i}{z+1}$ is purely imaginary,
So, $\operatorname{Re}\left(\frac{z-i}{z+1}\right)=0 \Rightarrow x^2+y^2-x-y=0$,
and it is a circle with centre $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$, excluding the points $(1,0)$ and $(0,1)$.