Solving $e^z = 1 $ in complex plane I am solving the equation $e^z = 1 $ in $\mathbb{C}$. The book says, other than $z = 0$, $z = 2 \pi k i$ for $ k \in \mathbb{Z}$ is also the solution. It explains the solution by saying that $e^z$ is periodic function so that
$1 = e^z = e^{2\pi k i}$
However I want to know how the identity is derived to solve other cases such as $e^z = 2 $.

You want to use the fact that $e^a = e^b$ if and only if $b-a = 2\pi ki$ for some integer $k$.
So as long as you can find some value of $z$ with $e^z=2$, this lets you find all such values.
For this example, you certainly know that $2 = e^{\ln 2}$ (where "$\ln$" is the ordinary real-valued function of a positive real variable), so
$$2 = e^z\tag{given equation to solve}$$
$$\iff e^{\ln 2} = e^z\tag{because $2=e^{\ln 2}$}$$
$$\iff z - \ln 2 = 2\pi ki\tag{rule I cited}$$
$$\iff z = \ln 2 + 2\pi ki\tag{solve for $z$}$$
for some integer $k$.

Now why is the rule true?
$e^a = e^b$ is equivalent to $e^{b-a} = 1$ (multiply both sides by $e^{-b}$).

So the rule is really the same as $e^w= 1$ if and only if $w = 2\pi k i$ for some integer $k$.
To see why this is true, remember that for real $u$ and $v$, we have $e^{u + iv} = e^u(\cos v + i\sin v)$.
This is $1$ precisely when $e^u=1$ (so $u=0$) and $(\cos v,\sin v) = (1,0)$ (so $v = 2\pi k$).
In other words, when $u+iv = 0 + 2\pi k i$, as required.
