If the amplitude of $(Z-2)$ is $\frac{\pi}{2}$,then the locus of $Z$ is:

$A$ function $f$ is defined on the complex numbers by $f(z) = (a + ib)z$,where $a, b \in \mathbb{R}^+$. This function has the property that the $f$-image of any point in the complex plane is equidistant from that point and the origin. If $|a + bi| = 10$ and $b^2 = \frac{p}{q}$,where $p, q \in \mathbb{Z}$ and $\text{gcd}(p, q) = 1$,then $p + q$ is:

Let $C$ be the set of all complex numbers. Let $S_{1} = \{z \in C : |z-3-2i|^{2}=8\}$,$S_{2} = \{z \in C : \operatorname{Re}(z) \geq 5\}$,and $S_{3} = \{z \in C : |z-\bar{z}| \geq 8\}$. Then the number of elements in $S_{1} \cap S_{2} \cap S_{3}$ is equal to:

Match the statements in column-$I$ with those in column-$II$.
[Note: Here $z$ takes the values in the complex plane and $\operatorname{Im} z$ and $\operatorname{Re} z$ denote,respectively,the imaginary part and the real part of $z$]

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

If $z_1 = 10 + 6i$,$z_2 = 4 + 6i$ and $z$ is any complex number such that the argument of $\frac{z - z_1}{z - z_2}$ is $\frac{\pi}{4}$,then