I want to prove, that $ E(x) \neq 0 $ for $ E(x) := \cos x + i \cdot \sin x $ I came up with a proof, but I'm not sure whether my solution is right:
Say there is such an $ x $, that $ \cos(x) + i\sin x = 0 $.
(*)For this term to equal $0$, $ i $ must be annulled (I hope that's the right word) and therefor $ \sin x = 0 $.
Then: $ \sin x = 0 \implies x = k_1 \cdot \pi, k_1 \in \mathbb{Z} $
With $ i \cdot \sin x $ being $0$ we have $ \cos x = 0 $. Which would mean $ x = (2k_2 + 1) \cdot \frac{\pi}{2} $, $ k_2 \in \mathbb{Z} $ .
This would lead to $ (2k_2 + 1) \cdot \frac{\pi}{2} = k_1 \cdot \pi \iff k_1 = k_2 + \frac{1}{2}$, meaning that either $k_1$ or $k_2$ can't be an integer.
The theorems about $\sin x$ and $\cos x$ are given. The part I'm not so sure about is (*), so if it's legitimate to argue that $i$ must be annulled to get the real number $0$. I'm pretty new to complex numbers (and math in general), but as far as I know there would be no way to get the term equal 0, if $\sin x$ wouldn't equal 0. Is this correct?
If $a$ and $b$ are real numbers, then $a+bi = 0$ if and only if $a$ and $b$ are both $0$, so the part you mark (*) in your proof is valid.
In fact, a more general fact is true, and that is that if $a_1, a_2, b_1, b_2$ are all real numbers, then $a_1+b_1i = a_2 + b_2i \iff a_1=a_2\land b_1 = b_2$. In your case, set $a_2=b_2 = 0$, and you get the statement you need.