Show that $e^{\frac{\pi}{2}i }= i$ - Problem to understand a certain convergence

Question

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

$1=|e^{\frac{\pi}{2}i}|=|\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})|=|\sin(\frac{\pi}{2})|$

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And now the part that I don't understand

Because $\sin'(x)=\cos(x)>0,\forall_{x\in[0,\frac{\pi}{2})}$

$\Longrightarrow\sin(\frac{\pi}{2})=1$

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$\Longrightarrow e^{\frac{\pi}{2}i }= i$

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?

Answer 1

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 $$\sin\frac\pi2>0.$$

Answer 2

You know that $ \sin(x) \in \mathbb{R}$, so that $$\left\vert \sin \left( \frac{\pi}{2} \right) \right\vert=1 \Rightarrow \sin \left( \frac{\pi}{2} \right)= \pm1$$

If $\sin'(x)=\cos(x) >0$ for all $x \in [0,\frac{\pi}{2})$, then this means that the sinus function is strictly growing on $x \in [0,\frac{\pi}{2})$. With $\sin(0)=0$, with $\sin$ being an continuous function, then the only possibility is $$\sin \left( \frac{\pi}{2} \right)=+1$$

Hence $$e^{i\frac{\pi}{2}} = i$$

Edit If you want to proove that $\sin'(x)=\cos(x)$, so note that $$\frac{d}{dx}\left( e^{ix} \right) = ie^{ix} = i\cos(x) - sin(x)$$ Implies that $$ \cos'(x) + i\sin'(x) = -\sin(x) + i\cos(x)$$ Prooving that $\sin'(x)=\cos(x)$