If $e^{i\theta}=e^{i\varphi}$, then $\theta-\varphi=2k\pi$

Question

This is pretty easy I think but I am having a tough time trying to prove this in a satisfying way to me. I am trying to show that $$e^{i\theta}=e^{i\varphi} \Rightarrow \theta-\varphi=2k\pi,\, \text{ for some $k \in \mathbb{Z}$.}$$ I know $e^{i\theta}=e^{i\varphi}$ implies $$\cos(\theta)=\cos(\varphi) \text{ and }\sin(\theta)=\sin(\varphi).\qquad (1)$$ An argument that I thought of but want to avoid if I can, is arguing along the lines of saying: Say $\theta$ is in quadrant $I$, then necessarily, $\varphi$ is then in quadrant $I$ or quadrant $IV$. If $\varphi$ is in quadrant $I$, then we're done. If $\varphi$ is in quadrant $IV$ then $\sin(\varphi)<0$ but $\sin(\theta)>0$ (here assuming $\sin(\theta)>0$ to give idea behind reasoning), a contradiction so we'd be done. Then fitting this to the different cases or doing similar kinds of reasoning.

I was hoping to get some kind of algebraic equation from $(1)$ and from there deduce necessarily that $\varphi-\theta=2k\pi$.

Answer 1

Starting with $e^{i \theta}=e^{i \psi}$, multiply both sides by $e^{-i \psi}$ to get $e^{i \theta}e^{-i \psi}=1$. Using rules of exponents this means $e^{i( \theta - \psi)}=1$. Now applying Euler's identity to the exponent means $cos(\theta - \psi)+isin(\theta - \psi)=1$. Since the RHS has no imaginary component, it follows that $isin(\theta - \psi)=0$, which happens precisely when $\theta - \psi=k \pi$ (any integer multiple of $\pi$). This leaves $cos(\theta - \psi)=1$, which is true only when the argument inside cosine is an even multiple of $\pi$. Hence, $\theta - \psi = 2k \pi$.