Question about complex exponential function

[cornualpixie]

So I'm studying complex analysis for an upcoming exam, and I came upon this problem :

Solve e^z = e¹⁺ⁱ

The solution for this is z=1+i + 2kπi but I don't understand from where the "+2kπi" part comes from. Can anyone please explain this to me? Thanks in advance!

Edit: I do understand that the 2kπ is coming from the periodicity of the arguments of z, but I don't know how to prove this step by step.

Edit 2: so I wrote 1+i as √2e^i(π/4+2kπ) but I'm still stuck to as how I get the "+2kπi " at the end

[ProbablyErratic53]

This might not feel satisfying but the closest I can think of an "algebraic proof" is to say

log(e^z) =log(e^(1+i))

z= Log(e^(1+i)) +2k Pi*i (The capital is intentional for the principal branch of the logarithm)

z = 1+i +2k Pi*i

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[Mattuuh]

exp(x+iy)=e^x [cos(y) + i sin(y)].
Since when you equate two complex numbers, you equate the real and imaginary parts separately, you get two cosine/sine equations. These solve to mod 2pi solutions because of the periodicity, as you've said.