If $z$ lies on the circle $|z-1|=1$ then$\frac {z-2} z$ is a purely imaginary number If $z$ lies on the circle $|z-1|=1$ then $\frac {z-2} z$ is a purely imaginary number. This is what by book states. Did'nt understand why. Can someone help?
Actually I was thinking of a more geometrical approach to the problem, as pointed out by @did and the answer below. I also remembered the method that if $\bar z = -z$ then $z$ is purely imaginary. But can it also be done by visual geometry? Ok, I drew a circle centred at $1$ with radius $1$. The statement given implies $\frac {z-2} z$ equals any point on the semicircle. After that how do we prove its imaginary from there?

Let $z = x + iy$. Then $|z - 1|^2 = 1 \rightarrow (x-1)^2 + y^2 = 1$.
Then show that the real part of $\frac{z-2}{z}$ is $x(x - 2) + y^2$, which is zero, from above.
