PREVIOUS YEAR IIT-JEE QUESTION COMPLEX NUMBERS

$\boldsymbol{Ans:(2):}$\Rightarrow$ $z^3=\bar z.........................(1)$\\\\ Multiply both side by $z$ we get $z^3.z = z.\bar z\Leftrightarrow \underbrace{z^4}_{L.H.S}=\underbrace{|z|^2}_{R.H.S}$\\\\ Here R.H.S is a Real No. So L.H.S must be a Real quantity.\\\\ So $z$ is either Purely Real or Purely Imaginary.\\\\ \underline{\underline{Case:(1):}}$\Rightarrow$ If $z=a$.Where $a\in R$. Then put into equation....(1),We get\\\\ $a^3=a\Leftrightarrow a.(a^2-1)=0$. So $\boxed{a=0,\pm 1}$\\\\ So $\boxed{\boxed{z=0,\pm 1}}$\\\\ Similarly\\\\ \underline{\underline{Case:(2):}}$\Rightarrow$ If $z=i.a$.Where $a\in R$. Then put into equation....(1),We get\\\\ $i^3.a^3=-i.a\Leftrightarrow a.(a^2-1)=0$. So $\boxed{a=0,\pm 1}$\\\\ So $\boxed{\boxed{z=\pm i}}$\\\\ So all Complex no. that satisfy the equation is $\boxed{\boxed{z=0,\pm i,\pm 1}}$

$\boldsymbol{Ans:(1):}$\Rightarrow$ $4z^2+8|z^2|=8\Leftrightarrow z^2+2|z|^2=2.........................(1)$\\\\ $z^2 = 2(1-|z|^2).\Leftrightarrow \underbrace{z^2}_{L.H.S}=\underbrace{2.(1-|z|^2)}_{R.H.S}$\\\\ Here R.H.S is a Real No. So L.H.S must be a Real quantity.\\\\ So $z$ is either Purely Real or Purely Imaginary.\\\\ \underline{\underline{Case:(1):}}$\Rightarrow$ If $z=a$.Where $a\in R$. Then put into equation....(1),We get\\\\ $a^2+2a^2=2\Leftrightarrow 3a^2=2$. So $\boxed{a=\pm\sqrt{\frac{2}{3}}}$\\\\ So $\boxed{\boxed{z=\pm \sqrt{\frac{2}{3}}}}$\\\\ Similarly\\\\ \underline{\underline{Case:(2):}}$\Rightarrow$ If $z=i.a$.Where $a\in R$. Then put into equation....(1),We get\\\\ $i^2.a^2+2a^2=2\Leftrightarrow a^2=2$. So $\boxed{a=\pm\sqrt{2}}$\\\\ So $\boxed{z=\pm\sqrt{2}}$ not Satisfy the eqaution .......(1)\\\\ So all Complex no. that satisfy the equation is $\boxed{\boxed{z=\pm \sqrt{\frac{2}{3}}}}$

2) z^3 = \overline{z} \Rightarrow (\overline{z})^3 = z

So, z^9 = (\overline{z})^3 = z \Rightarrow z=0 \ \text{or} \ z^8=1

Hence the solutions are z=0 and the eighth roots of unity.

Thanks hsbhatt Sir,
Now I understand.