Let the locus of a point z in the Argand plan | vedclass

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(D) Let $z=x+iy$, then $z^2=(x+iy)^2 = x^2-y^2+i(2xy)$.For $C_1$, $\operatorname{Re}(z^2)=x^2-y^2=4$ $(i)$.For $C_2$, $\operatorname{Im}(z^2)=2xy=4 \R

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(D) Let $z=x+iy$, then $z^2=(x+iy)^2 = x^2-y^2+i(2xy)$.For $C_1$, $\operatorname{Re}(z^2)=x^2-y^2=4$ $(i)$.For $C_2$, $\operatorname{Im}(z^2)=2xy=4 \Rightarrow xy=2$ $(ii)$.From $(ii)$, $y=\frac{2}{x}$. Substituting into $(i)$:$x^2 - (\frac{2}{x})^2 = 4 \Rightarrow x^2 - \frac{4}{x^2} = 4$.Let $t=x^2$ (where $t > 0$): $t - \frac{4}{t} = 4 \Rightarrow t^2 - 4t - 4 = 0$.Using the quadratic formula, $t = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2} = 2 \pm 2\sqrt{2}$.Since $t=x^2 > 0$, we must have $t = 2+2\sqrt{2}$.Thus, $x^2 = 2+2\sqrt{2}$, which gives two real values for $x$ $(x = \pm \sqrt{2+2\sqrt{2}})$.For each $x$, $y = \frac{2}{x}$ gives a unique real value for $y$.Therefore, there are $2$ common points.

column-$I$	column-$II$
$(A)$ The set of points $z$ satisfying $\|z-i\|z\|\|=\|z+i\|z\|\|$ is contained in or equal to	$(p)$ an ellipse with eccentricity $\frac{4}{5}$
$(B)$ The set of points $z$ satisfying $\|z+4\|+\|z-4\|=10$ is contained in or equal to	$(q)$ the set of points $z$ satisfying $\operatorname{Im} z=0$
$(C)$ If $\|\omega\|=2$,then the set of points $z=\omega-1/\omega$ is contained in or equal to	$(r)$ the set of points $z$ satisfying $\|\operatorname{Im} z\| \leq 1$
$(D)$ If $\|\omega\|=1$,then the set of points $z=\omega+1/\omega$ is contained in or equal to	$(s)$ the set of points $z$ satisfying $\|\operatorname{Re} z\| \leq 1$
	$(t)$ the set of points $z$ satisfying $\|z\| \leq 3$

Let the locus of a point $z$ in the Argand plane satisfying the condition $\operatorname{Re}(z^2)=4$ be $C_1$ and the locus of $z$ satisfying the condition $\operatorname{Im}(z^2)=4$ be $C_2$. Then the number of common points of the two curves $C_1$ and $C_2$ are

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