Let $z$ and $w$ be two non-zero complex numbers such that $|z| = |w|$ and $arg(z) + arg(w) = \pi$. Then $z$ is equal to:

For $a \in \mathbb{C}$, let $A = \{z \in \mathbb{C} : \operatorname{Re}(a + \bar{z}) > \operatorname{Im}(\bar{a} + z)\}$ and $B = \{z \in \mathbb{C} : \operatorname{Re}(a + \bar{z}) < \operatorname{Im}(\bar{a} + z)\}$. Then among the two statements:
$(S1) : \text{If } \operatorname{Re}(a), \operatorname{Im}(a) > 0, \text{ then the set } A \text{ contains all the real numbers.}$
$(S2) : \text{If } \operatorname{Re}(a), \operatorname{Im}(a) < 0, \text{ then the set } B \text{ contains all the real numbers.}$

If the imaginary part of $\frac{2z + 1}{iz + 1}$ is $-2$,then the locus of the point representing $z$ in the complex plane is

Let $z=x+iy$ and a point $P$ represent $z$ in the Argand plane. If the real part of $\frac{z-1}{z+i}$ is $1$,then a point that lies on the locus of $P$ is

If $z_{1}, z_{2}$ are complex numbers such that $\operatorname{Re}(z_{1})=|z_{1}-1|$, $\operatorname{Re}(z_{2})=|z_{2}-1|$ and $\arg(z_{1}-z_{2})=\frac{\pi}{6}$, then $\operatorname{Im}(z_{1}+z_{2})$ is equal to

If $|8 + z| + |z - 8| = 16$,where $z$ is a complex number,then the point $z$ will lie on