(d) Let ${z_1} = {r_1}$$(\cos {\theta _1} + i\sin {\theta _1})$,${z_2} = {r_2}$ $(\cos {\theta _2} + i\sin {\theta _2})$
$\therefore $$|{z_1} + {z_2}| = [{({r_1}\cos {\theta _1} + {r_2}\cos {\theta _2})^2}$
$ + {({r_2}\sin {\theta _1} + {r_2}\sin {\theta _2})^2}{]^{1/2}}$
$ = {[r_1^2 + r_2^2 + 2{r_1}{r_2}\cos ({\theta _1} – {\theta _2})]^{1/2}} = {[{({r_1} + {r_2})^2}]^{1/2}}$

$|{z_1} + {z_2}|\, = \,|{z_1}| + |{z_2}|$
Therefore $\cos ({\theta _1} – {\theta _2}) = 1 \Rightarrow {\theta _1} – {\theta _2} = 0 \Rightarrow {\theta _1} = {\theta _2}$
Thus arg $({z_1}) – arg({z_2}) = 0$.
Trick : $|{z_1} + {z_2}|\, = \,|{z_1}| + |{z_2}| \Rightarrow {z_1},{z_2}$ lies on same straight line.
$\therefore arg\,{z_1} = arg\,{z_2} \Rightarrow arg\,{z_1} – arg\,{z_2} = 0$.