Explanation

(b)
We know that, is
$\left| z_{1} + z_{2} \right| = \left| z_{1} \right| + |z_{2}|$, then $\arg{\left( z_{1} \right) = \arg\left( z_{2} \right)}$
$\because\ \ \ \ \ \left| z^{2} + ( - 1) \right| = \left| z^{2} \right| + | - 1|$
$\Rightarrow \ \ \ \ \ \arg{\left( z^{2} \right) = \arg( - 1)}$
$\Rightarrow \ \ \ \ 2\arg{(z) = \pi}$ $\lbrack\because\ \ \arg{( - 1) = \pi}\rbrack$
$\Rightarrow \ \arg{(z) = \frac{\pi}{2}}$
$\Rightarrow \ \ z$ lies on $y$-axis (imaginary axis).