For a complex number $z$, let $\operatorname{Re}(z)$ denote the real part of $z$. Let $S$ be the set

Question

For a complex number $z$, let $\operatorname{Re}(z)$ denote the real part of $z$. Let $S$ be the set of all complex numbers $z$ satisfying $z^{4}-|z|^{4}=4 i z^{2}$, where $i=\sqrt{-1}$. Then the minimum possible value of $\left|z_{1}-z_{2}\right|^{2}$, where $z_{1}, z_{2} \in S$ with $\operatorname{Re}\left(z_{1}\right)>0$ and $\operatorname{Re}\left(z_{2}\right)<0$, is

Professor PC Thomas Classes · Answer

(8) : $z^{4}-|z|^{4}=4 i z^{2}$\ $\Rightarrow z^{4}-(z \bar{z})^{2}=4 i z^{2} \Rightarrow z^{2}\left|z^{2}-\bar{z}^{2}ight|=4 i z^{2}$\ $\Rightarrow \quad z^{2}=0$ or $z^{2}-\bar{z}^{2}=4 i$\ $\Rightarrow \quad(z+\bar{z})(z-\bar{z})=4 i \Rightarrow 2 x(2 i y)=4 i$\ $	herefore \quad x y=1$\ The minimum distance occur along with normal. Let $(t, 1 / t)$ be a point on $C_{1}$.\ $	herefore \frac{d y}{d x}=\frac{-1}{t^{2}}$\ Slope of normal $=t^{2}(t>0)$\ Similarly, let $(s, 1 / s)$ be a point on $C_{2},(s<0)$.\ $	herefore \quad$ Slope of normal $=s^{2}$\ Slope of $A B=\frac{\frac{1}{t}-\frac{1}{s}}{t-s}=-\frac{1}{t s}$\ From $t^{2}=s^{2}=-\frac{1}{t s}$, we get $t=1, s=-1$\ Hence, the shortest distance\ $=\sqrt{(1+1)^{2}+(1+1)^{2}}=\sqrt{8}$\ $	herefore\left|z_{1}-z_{2}ight|^{2}=8$\