For a complex number z , let 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=√(-1) . Then the minimum possible value of z 1 -z 2 2 , where z 1 , z 2 in S with Re ( z 1 ) 0 and Re ( z 2 )<0 , is \ \ \ \

Mathematics · Complex Numbers

JEE Advanced 2020 — Paper 2 — Question 34

For a complex number z , let $\operatorname{Re}(\mathrm{z})$ 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 $\operatorname{Re}\left(\mathrm{z}_{1}\right)>0$ and $\operatorname{Re}\left(\mathrm{z}_{2}\right)<0$ , is $\_\_\_\_$

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

\begin{array}{ll} & z^{4}-|z|^{4}=4 i z^{2} \\& z^{4}-z^{2} \cdot \bar{z}^{2}=4 i z^{2} \Rightarrow \\& z^{2}=0 \text { or } z^{2}-\bar{z}^{2}=4 i\\ \text { Let } z=x+i y \Rightarrow \\& (z+\bar{z})(z-\bar{z})=4 i \\& \Rightarrow 2 x \cdot 2 i y=4 i \Rightarrow & x y=1 \end{array}

$\mathrm{z}_{1}$ lies on $\mathrm{xy}=1$ in first quadrant and $\mathrm{z}_{2}$ lies on $\mathrm{xy}=1$ in third quadrant.

$\Rightarrow\left|z_{1}-z_{2}\right|^{2}$ is minimum when $z_{1} \equiv(1,1)$ and $z_{2}=(-1,-1)$

$\Rightarrow\left|z_{1}-z_{2}\right|^{2}=8$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 2
Subject: Mathematics
Chapter: Complex Numbers
Topic: Geometry of Complex Numbers