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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

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  • $\begingroup$ If $i$ is the imaginary number, it is normal. Add your syntax to the post or give a link. $\endgroup$ – Claude Leibovici Sep 28 at 12:03
  • $\begingroup$ I already edited it with the code line and the link, but what do you mean it's normal? And yes, $i=\sqrt{-1}$ $\endgroup$ – TheWonkaBro Sep 28 at 12:08
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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 $f(x+2\pi k)$.

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  • $\begingroup$ what do you mean shifted by $2k\pi$? won't $e^{2ik\pi}$ always be equal one? $\endgroup$ – TheWonkaBro Sep 28 at 12:23
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    $\begingroup$ That's right. However, after dividing the exponent by $i$ you'll get different answers $(e^{ix})^{\frac{1}{i}}=e^x$, $(e^{i(x+2\pi k)})^{\frac{1}{i}}=e^{x+2\pi k}$ $\endgroup$ – cansomeonehelpmeout Sep 28 at 12:26
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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:

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

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