We have defined $e^{ix}=\cos x+ i \sin x$ where $x\in\mathbb{R}$.
I also have proved earlier that $\cos(\frac{\pi}{2})=0$ and know that $\sin(0)=0$
Now I want to prove that
$e^{\frac{\pi}{2}i }= i$
I know that
$|e^{xi }|= 1,\forall_{x\in\mathbb{R}}$
Therefore
$1=|e^{\frac{\pi}{2}i}|=|\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})|=|\sin(\frac{\pi}{2})|$
And now the part that I don't understand
Because $\sin'(x)=\cos(x)>0,\forall_{x\in[0,\frac{\pi}{2})}$
$\Longrightarrow\sin(\frac{\pi}{2})=1$
$\Longrightarrow e^{\frac{\pi}{2}i }= i$
Please explain me which Theorem we use here. And how this Situation can be generalised, so if I Encounter something similiar I directly know what to do
Edit:
Can it be shown constructively, i.e. with the mean-value-theorem for example?
Best Answer
Essentially we need only to prove $\sin\frac\pi2>0$. The answer will use the mean value theorem (MVT) as asked in question.
The definition of $e^z,\sin z, \cos z$ implies that the functions are continuous and differentiable. Particularly: $$ \forall z\in\mathbb C: \sin'z=\cos z.\tag1 $$
As $\frac\pi2$ is by definition (see discussion in comments) the least positive root of $\cos x$ and $\cos(0)=1>0$ we have from the continuity of the function: $$ \forall x\in\left(0,\frac\pi2\right): \cos x>0. $$
Together with (1) this implies: $$ \forall x\in\left(0,\frac\pi2\right): \sin' x>0.\tag2 $$
Now by MVT there exists such a point $0<c<\frac\pi2$ that $$ \sin'(c)=\frac{\sin\frac\pi2-\sin0}{\frac\pi2-0} \implies \sin\frac\pi2=\frac\pi2\sin'(c). $$ As both $\sin'(c)$ and $\frac\pi2$ are positive, we obtain $$\sin\frac\pi2>0.$$