Prove $e^i=i^{\frac{2}{\pi}}$ true or false?

Question

Assume $~e^{i~x}=\cos(x)~+~i~\sin(x)~$ is questionable, can't be used.

Answer 1

$e^i$, value of the exponential function, has one value, $\cos 1 + i \sin 1$.

But $i^{2/\pi}$ has many values, since that is how the complex exponent $u^v$ works.

$$ i^{2/\pi} = \exp\left(\frac{2}{\pi}\log(i)\right) $$ and multi-valued $\log(i)$ has values $$ \frac{\pi i}{2} + 2\pi k i,\qquad k \in \mathbb Z $$ so $i^{2/\pi}$ has values $$ \exp\left(\frac{2}{\pi}\;i\;\left(\frac{\pi + 2\pi k}{2}\right)\right) = \exp\left(i\;\left(1 + 2 k\right)\right) = \cos(1+2k) + i \sin(1+2k),\qquad k \in \mathbb Z $$ Yes. One of the values is, indeed, $e^i = \cos 1 + i \sin 1$. But it is best not to say $$ e^i = i^{2/\pi} $$ because of the other values.

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Suppose we interpret $e^i$ as a complex exponent. Then it also has many values. $$ \exp(i\log(e)) $$ and complex $\log(e)$ has values $$ 1+2\pi i k,\qquad k \in \mathbb Z $$ so all values are $$ \exp(i(1+2\pi i k)) = \exp(i-2\pi k) = e^{-2\pi k}\big(\cos 1 + i \sin 1\big) $$ All values of $e^i$ except one are different from values of $i^{2/\pi}$. Indeed, the values of $i^{2/\pi}$ all have modulus $1$, while the values of $e^i$ have modulus $e^{-2\pi k}$, and these are all $\ne 1$ except for the one value with $k=0$.