JEE Main 2022 — Complex Number Question with Solution

Given,

$z^2=\overline{z}·2^{1-\left|z\right|} \cdots \left(1\right)$

On taking modulus both side we get,

$\Rightarrow {\left|z\right|}^2=\left|\overline{z}\right|·2^{1-\left|z\right|}$

$\Rightarrow \left|z\right|=2^{1-\left|z\right|}, \because b\ne 0\Rightarrow \left|z\right|\ne 0$

So comparing both side we get $\left|z\right|=1 \cdots \left(2\right)$

Now putting $z=a+ib$ then $\sqrt{a^2+b^2}=1 \cdots \left(3\right)$

Now again from equation $\left(1\right)$, equation $\left(2\right)$, equation $\left(3\right)$ we get:

$a^2-b^2+i2ab=\left(a-ib\right)2^0=a-ib$

Now on comaparing imagenary and real part we get,

$\therefore a^2-b^2=a$ and $2ab=-b$

Now solving we get, $a=-\frac{1}{2}$ and $b=\pm \frac{\sqrt{3}}{2}$

So, $z=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ or $=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$

Now solving $z^n={\left(z+1\right)}^n\Rightarrow {\left(\frac{z+1}{z}\right)}^n=1$

$\Rightarrow {\left(1+\frac{1}{z}\right)}^n=1$

$\Rightarrow {\left(\frac{1+\sqrt{3}i}{2}\right)}^n=1$

$\Rightarrow {\left(-{\omega }^2\right)}^n=1$, then minimum value of $n$ is $6$