If $z$ lies on the circle $|z-1|=1, $ then the what is the value of $\frac{z-2}{z}$ ($z\ne 0 \ \text{and} \ 2$ ) is?
What I am aware of : the equation $|z-1|=1$ represent a circle with centre $(1,0)$ and radius $1$.
Also $$|z-1|=1 \implies (z-1)^2=1 \\ \implies z^2-2z+1=1 \implies z(z-2) =0$$
$\implies z = 0; z =2 $
I think this is wrong.. please suggest thanks..
$(a)\implies z-2=0\implies z=2\implies|z-1|=1$: on the circle.
$(b)\implies z-2=2z\implies z=-2\implies|z-1|=3$: not on the circle.
$(c)\implies z-2=-z\implies z=1\implies|z-1|=0$: not on the circle.
Exercise: The image of the circle of equation $|z-1|=1$ by the function $z\mapsto(z-1)/z$ is the line of equation $\Re(z)=0$.
Note that the function $$ f(z) = \frac{z-2}{z} $$ is holomorphic except at $z=0$. Hence, there is no way that $f$ can be constant on the circle $|z-1| = 1$ (that would violate the uniqueness theorem for holomorphic functions).
Let us set $z=x+iy$ where $x,y$ are real
So, $1=|z-1|=|x+iy-1|=\sqrt{(x-1)^2+y^2}\implies (x-1)^2+y^2=1$
For parametrization, we can set $x-1=\cos2\theta,y=\sin2\theta$
$$\implies\frac{z-2}z=\frac{1+\cos2\theta+i\sin2\theta-2}{1+\cos2\theta+i\sin2\theta}=\frac{i2\sin\theta\cos\theta-2\sin^2\theta}{2\cos^2\theta+i2\sin\theta\cos\theta}=\frac{2i\sin\theta(\cos\theta+i\sin\theta)}{2\cos\theta(\cos\theta+i\sin\theta)}=i\tan\theta$$ which is purely imaginary unless $\theta=n\pi$ (where $n$ is any integer), whence it will be $0$
The fact that a complex number $w$ has modulus equal to $1$, $\vert w\vert = 1$, doesn't imply that $w^2 = 1$, as you used in your first implication, with $w = z-1$. For instance: $\vert i \vert = 1$ and $i^2 = -1$.
