Equality of complex numbers in general Suppose $z_1 = r_1(\cos \theta_1 + i\sin \theta_1)$ and $z_2 = r_2(\cos \theta_2 + i\sin \theta_2)$. Prove that $z_1 = z_2$ $\iff$ $r_1 = r_2 , \theta_1 = \theta_2 + 2k\pi$ .
My try: The proof for the converse statement is obvious but I don't know how to prove the forward statement. I tried to use "Two complex numbers $a+bi$ and $c+di$ are equal iff $a=b$ and $c=d$, where $a,b,c,d$ are real numbers." but it didn't work.
 A: Note that if $z=r(\cos\theta+i\sin\theta)(r\ge0)$ then $|z|=r$. In fact, $|z|^2=(r\cos\theta)^2+(r\sin\theta)^2=r^2$, then $|z|=r$.
Here I assume $r_1,r_2>0$. Otherwise $z=0=0\cdot(\cos\theta+i\sin\theta)$ holds for every $\theta\in\mathbb R$, in which case the claim in OP's problem is wrong.
$z_1=z_2$ implies that $r_1=|z_1|=|z_2|=r_2$, so $\cos\theta_1=\cos\theta_2$ and $\sin\theta_1=\sin\theta_2$, which means $\theta_1=\theta_2+2k\pi$.
A: \begin{align}z_1=z_2&\implies r_1\cos\theta_1+ir_1\sin\theta_1=r_2\cos\theta_2+ir_2\sin\theta_2\\&\implies\begin{cases}r_1\cos\theta_1=r_2\cos\theta_2\\r_1\sin\theta_1=r_2\sin\theta_2\end{cases}\\&\implies \tan\theta_1=\tan\theta_2\quad (r_1,r_2>0\,\land\,\cos\theta_1,\cos\theta_2\ne0 \,\,{}^{**})\\&\implies\theta_1=\theta_2+K\pi\quad(K\in\Bbb Z)\\&\implies r_2\cos\theta_2=r_1\cos(\theta_2+K\pi)=r_1\cos\theta_2\cos K\pi-r_1\sin\theta_2\sin K\pi\\&\implies r_2\cos\theta_2=r_1\cos\theta_2\cos K\pi\\&\implies r_2=r_1\cos K\pi\quad(\cos\theta_2\ne0)\\&\implies r_2=r_1\quad(\cos K\pi=\pm1;\,r_1,r_2>0\implies\cos K\pi=1\implies K\equiv0\pmod2)\\&\implies r_1=r_2\,\land\, \theta_1=\theta_2+2k\pi\quad(k\in\Bbb Z)\end{align}
${}^{**}$

 \begin{align}\cos\theta_1,\cos\theta_2=0&\implies\theta_1,\theta_2=\frac\pi2+K\pi\quad(K\in\Bbb Z)\\&\implies\sin\theta_1,\sin\theta_2=\pm1\\\cos\theta_1,\cos\theta_2=0&\implies r_1(0+i\sin\theta_1)=r_2(0+i\sin\theta_2)\\&\implies r_1\sin\theta_1=r_2\sin\theta_2\\r_1,r_2>0&\implies\sin\theta_1=\sin\theta_2=1\\&\implies\theta_1=\theta_2+2k\pi\quad(k\in\Bbb Z)\end{align}

A: I think you meant that $r_i\geq0$ and $\theta_i\in\mathbb R,$ otherwise your statement is wrong.
Now, by your work we obtain:
$$(r_1\cos\theta_1)^2+(r_1\sin\theta_1)^2=(r_2\cos\theta_2)^2+(r_2\sin\theta_1)^2$$ or
$$r_1^2=r_2^2,$$ which gives $$r_1=r_2.$$
Can you end it now?
A: If $\cos\theta_1=\cos\theta_2$  and   $\sin\theta_1=\sin\theta_2$ 
Using Prosthaphaeresis Formulas,
$$0=\cos\theta_1-\cos\theta_2=-2\sin\dfrac{\theta_1-\theta_2}2\sin\dfrac{\theta_1+\theta_2}2$$
$$0=\sin\theta_1-\sin\theta_2=2\sin\dfrac{\theta_1-\theta_2}2\cos\dfrac{\theta_1+\theta_2}2$$
If  $\sin\dfrac{\theta_1-\theta_2}2=0, \dfrac{\theta_1-\theta_2}2$ must be multiple of $\pi$ and we are done.
Else $\sin\dfrac{\theta_1+\theta_2}2=\cos\dfrac{\theta_1+\theta_2}2=0$ which is untenable as $$\sin^2\dfrac{\theta_1+\theta_2}2+\cos^2\dfrac{\theta_1+\theta_2}2=1$$
A: Since $z_1^\ast z_1=z_2^\ast z_2$, $r_1^2=r_2^2$ so $r_1=r_2$. Thus $\cos\theta+i\sin\theta_1=\cos\theta_2+i\sin\theta_2$. Dividing by the unit complex number on the right-hand side, $\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)=1$, i.e. $\cos(\theta_1-\theta_2)=1,\,\sin(\theta_1-\theta_2)=0$. Hence $2\pi|\theta_1-\theta_2$.
