Let the locus of a point $\mathrm{z}$ in the Argand plane satisfying the condition…
Let the locus of a point $\mathrm{z}$ in the Argand plane satisfying the condition $\operatorname{Re}\left(\mathrm{z}^2\right)=4$ be $\mathrm{C}_1$ and the locus of $z$ satisfying the condition $\operatorname{Im}\left(\mathrm{z}^2\right)=4$ be $\mathrm{C}_2$. Then the number of common points of the two curves $\mathrm{C}_1$ and $\mathrm{C}_2$ are
$0$
$3$
$1$
$2$
Solution
Let $z=x+i y \Rightarrow z^2=x^2-y^2+i 2 x y$
$\begin{aligned} & \operatorname{Re}\left(z^2\right)=x^2-y^2=4 ..(i)\\ & \operatorname{Im}\left(z^2\right)=2 x y=4 \Rightarrow x y=2 ...(ii)\end{aligned}$
$\begin{aligned} & \text { from } \mathrm{eq}^{\mathrm{n}} \text { (i) and (ii) } \mathrm{x}^2-\frac{4}{\mathrm{x}^2}=4 \\ & \Rightarrow \mathrm{x}^4-9 \mathrm{x}^2-9=0\end{aligned}$
$\Rightarrow \quad x^2=\frac{4 \pm \sqrt{16+16}}{2}=2 \pm 2 \sqrt{2}$
$\begin{aligned} & \Rightarrow \quad x^2=2(1 \pm \sqrt{2}) \\ & \because \quad x^2=2(1-\sqrt{2}) < 0 \text { (Not possible). }\end{aligned}$
$\Rightarrow x^2=2(1+\sqrt{2}) \Rightarrow x= \pm[2(1+\sqrt{2})]^{\frac{1}{2}}$
So, $y= \pm \frac{2}{[2(1+\sqrt{2})]^{\frac{1}{2}}}$
$\therefore$ We get 2 common points.