JEE Advanced 2022 Paper 1 · Q05 · Complex Numbers

Question

Let $\bar{z}$ denote the complex conjugate of a complex number $z$ and let $i=\sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\bar{z}-z^{2}=i(\bar{z}+z^{2})$ is _____________.

SolveKar AI · Accepted Answer

Rewrite as $\bar{z}(1-i)=z^{2}(1+i)\Rightarrow \bar{z}=z^{2}\cdot\dfrac{1+i}{1-i}=z^{2}\cdot i$. So $\bar{z}=iz^{2}$. Taking moduli: $|z|=|z|^{2}$, hence $|z|=0$ or $|z|=1$. Case 1: $z=0$ is a root. Case 2: $|z|=1$, so $\bar{z}=1/z$. Then $1/z=iz^{2}\Rightarrow z^{3}=1/i=-i=e^{-i\pi/2}$. The cube roots are $z=e^{-i\pi/6}, e^{i\pi/2}, e^{i 7\pi/6}$; all distinct unit-modulus numbers. Total distinct roots: $1+3=4$.

JEE Advanced 2022 Paper 1 · Q05 · Complex Numbers

Solution