IIT-JEE 1999Question 1
For complex numbers $z$ and $w$, prove that $|z|^{2}w-|w|^{2}z=z-w$ if and only if $z=w$ or $z\bar w=1$.
Complex Numbers
Answerproved
Solution
If $z=w$ both sides are $0$. Conversely the equation gives $(|z|^2+1)w=(|w|^2+1)z$, so $w=\lambda z$ with $\lambda>0$ real, and $(|z|^2+1)\lambda=\lambda^2|z|^2+1$, i.e. $(\lambda-1)(\lambda|z|^2-1)=0$. So $\lambda=1$ ($z=w$) or $\lambda|z|^2=1$, i.e. $z\bar w=\lambda|z|^2=1$. The converse for $z\bar w=1$ follows by reversing: $w=z/|z|^2$ satisfies the equation.