I turned this proof in for homework and think it may have several flaws. In the forward direction of the implication is it okay to divide by $r_2$ and assume $r_2$ is nonzero?Also in the backwards direction is it okay to divide by $z_2$?
Prove $\text{arg}{z_1}=\text{arg}z_2$ if and only if $z_1=cz_2$ where $c>0$
Attempt:
$\text{let} z_1=r_1(\cos\theta+i\sin\theta)$, $z_2=r_2(\cos\theta+i\sin\theta)$
Thus $\frac{z_2}{r_2}=\cos\theta+i\sin\theta \implies z_1=\frac{z_2r_1}{r_2}$
Let $c=\frac{r_1}{r_2}$ then $z_1=cz_2$
Conversely suppose $z_1=cz_2 \implies c=\frac{z_1}{z_2}$
$\implies \text{arg}\frac{z_1}{z_2}=\text{arg}c \implies \text{arg}z_1-\text{arg}z_2=0 \implies \text{arg}z_1=\text{arg}z_2$