JEE Main 2022 — Mathematics Algebra

Given,

$${z}^{2}=\bar{z}\cdot {2}^{1-|z|}\cdots (1)$$z2=zˉ⋅21−∣z∣⋯(1)

On taking modulus both side we get,

$$\Rightarrow {|z|}^{2}=|z|\cdot {2}^{1-|z|}$$⇒∣z∣2=∣z∣⋅21−∣z∣

$$\Rightarrow |z|={2}^{1-|z|},\because b\neq 0\Rightarrow |z|\neq 0$$⇒∣z∣=21−∣z∣,∵b=0⇒∣z∣=0

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

Now putting $z=a+ib$z=a+ib then $\sqrt{{a}^{2}+{b}^{2}}=1\cdots (3)$a2+b2​=1⋯(3)

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

$${a}^{2}-{b}^{2}+i2ab=(a-ib){2}^{0}=a-ib$$a2−b2+i2ab=(a−ib)20=a−ib

Now on comaparing imagenary and real part we get,

$\therefore {a}^{2}-{b}^{2}=a$∴a2−b2=a and $2ab=-b$2ab=−b

Now solving we get, $a=-\frac{1}{2}$a=−21​ and $b=\pm \frac{\sqrt{3}}{2}$b=±23​​

So, $z=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$z=−21​+23​​i or $=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$=−21​−23​​i

Now solving ${z}^{n}={(z+1)}^{n}\Rightarrow {(\frac{z+1}{z})}^{n}=1$zn=(z+1)n⇒(zz+1​)n=1

$$\Rightarrow {(1+\frac{1}{z})}^{n}=1$$⇒(1+z1​)n=1

$$\Rightarrow {(\frac{1+\sqrt{3}i}{2})}^{n}=1$$⇒(21+3​i​)n=1

$\Rightarrow {(-{\omega }^{2})}^{n}=1$⇒(−ω2)n=1, then minimum value of $n$n is $6$6