How can this Complex problem?

Question

The problem is $e^z=i+\sqrt{3}$.

I put $z= x+yi$ ,then $i+\sqrt{3}$ is $2(\cos\left(\frac \pi 6\right)+\sin\left(\frac \pi 6\right)\cdot i)$.

So $e^x$ is $2$ and $y$ is $\frac \pi 6+2k\pi$. ($k$ is integer)

Is this right answer?

Answer 1

If you have the following equation: $$e^{x+yi}=1+\sqrt 3$$ Then, yes, you can infer $e^x=2$ and $y=\frac \pi 6+2\pi k$ for $k \in \Bbb{Z}$. Your reasoning is correct.

The answer is more complete if you solve the $e^x = 2$: namely $x = \ln 2$, and so the total answer is $z = \ln 2 + i(\frac{\pi}{6} + 2\pi k)$ for $k \in \mathbb{Z}$.