If $|z-1|=1$, where $z$ is a point on the argand plane, show that $\dfrac{z-2}{z}=i\cdot\tan(\arg z)$, where $i=\sqrt{-1}$ If $|z-1|=1$, where $z$ is a point on the argand plane, show that $\dfrac{z-2}{z}=i\cdot\tan(\arg z)$, where $i=\sqrt{-1}$
M y Approach: Let $z=x+iy$
$|z-1|^2=1\implies (z-1)(\bar z-1)=1\implies x^2+y^2-2x=0$
Now, $\dfrac{z-2}{z}=\dfrac{x+iy-2}{x+iy}=\dfrac{(x^2+y^2-2x+2iy)}{x^2+y^2}$
$\implies$ $\dfrac{z-2}{z}=i\cdot \dfrac{2y}{x^2+y^2}$
Now do I prove that $\dfrac{2y}{x^2+y^2}=\tan(\arg z)$
I know other methods to solve this problem. I Just want to know flaw in my approach.

You've already got $\dfrac{z-2}{z}=i\cdot \dfrac{2y}{x^2+y^2}$ (for $z\not=0$).
Now you can use $x^2+y^2=2x$ again to have $\dfrac{z-2}{z}=i\cdot \dfrac{2y}{2x}$.
