solving $cisX=cisY$ equations

Question

Part of a solution I came across with solving the following equation: $cis(n\frac{\pi}{3})=1$. I converted it to be $cis(n\frac{\pi}{3})=cis0$.

After trying to insert some values, I understood that the solution is $n\frac{\pi}{3}=2\pi k$ when $k\in\mathbb{Z}$ meaning $n=6k$.

Is it true to say that the solution of $cisX=cisY$ is $X=Y+2\pi k$ when $k\in\mathbb{Z}$?

Answer 1

Yes, it is. Two complex numbers are equal if and only if their real and imaginary parts coincide. Therefore, the equality

$\cos(X) + i \sin(X) = \cos(Y) + i \sin(Y)$

holds if and only if

$\cos(X) = \cos(Y)$ and $\sin(X) = \sin(Y)$.

Can you solve from here?

Answer 2

Your answer is completely correct but the sine function is zero twice in a period.

$$\sin(X)=0$$ $$X=0+k\pi=k\pi \ \ \ \ \ k∈Z$$ so $$X=3k \ \ \ \text{and not} \ \ \ 6k$$