Shouldn't the ith root of the complex exponential function be equal to the natural exponential function?

Question

When i plug $\sqrt[i]{e^{ix}}$ into WolframAlpha it shows a rather weird function considering it should show the natural exponential function, since $\sqrt[i]{e^{ix}}=(e^{ix})^{\frac{1}{i}}=e^{\frac{ix}{i}}=e^{x}$ Is this a bug or am I wrong? Thanks in advance

here's the link to the function in WolframAlpha:

https://www.wolframalpha.com/input/?i=%5Csqrt%5Bi%5D%7Be%5E%7Bix%7D%7D

And the code is \sqrt[i]{e^{ix}} if needed

Answer 1

Notice that $e^{ix}=e^{i(x+2\pi k)}$ for $k\in\Bbb Z$. Therefore you'll not only get $e^x$, but also the graph of $e^x$ shifted by $2\pi k$ for every $k\in\Bbb Z$.

That is $f(x)=\sqrt[i]{e^{ix}}$ satisfy $f^i=e^{ix}$, but so do $f(x+2\pi k)$.

Answer 2

The complex logarithm is multivalued,

$$\log z=\log|z|+i\angle z+2ik\pi$$

and if you want to get a function, you need to choose a "branch", i.e. a value of the integer $k$, which can be a function of $z$.

This has no impact on the antilogarithm,

$$e^{\log z}=e^{\log|z|+i\angle z+2ik\pi}=e^{\log|z|+i\angle z}(e^{2i\pi})^k=e^{\log|z|+i\angle z}\,1^k$$ still holds whatever $k$, but it makes a difference for powers:

$$z^w=e^{w\log z}=e^{w(\log|z|+i\angle z+2ik\pi)}=e^{w(\log|z|+i\angle z\pi)}(e^{2i\pi w})^k.$$