Does Euler's Formula prove all imaginary numbers equal $0$?

Question

Euler's formula is

$$ e^{i \theta } = \cos \theta + i \sin \theta $$

If we take $\theta$ to be $2 \pi$ we get

$$e^{2i \pi} = \cos (2 \pi ) + i \sin (2 \pi ) $$

Which simplifies to

$$e^{2i \pi} = 1$$

But we also know

$$e^0 = 1$$

Therefore, can we not say

$$e^0 = e^{2i \pi}$$

and thus,

$$0 = 2i \pi$$

Which leads to being a big problem for all imaginary numbers.

Example with $25i$:

$$\begin{align} 25i &= \frac{25 (2 \pi) i}{2 \pi} \\ 25i &= \frac{25 * 0}{2 \pi} \\ 25i &= 0 \\ \end{align}$$

Answer 1

You cannot say $e^{0}=e^{2\pi i}$ implies $0=2\pi i$.

The map $z\mapsto e^{z}$ is not one-to-one in the complex numbers.

Answer 2

The correct conclusion of $$e^a=e^b$$ is not that $a=b$, but rather that $a=b+2k\pi i$ for some $k\in\mathbb Z$.