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Given a positive number $x>0$ and a complex number $y\in\mathbb{C}$, is it justified to write the following?

$$(-x)^y=(e^{i\pi+\log(x)})^y=e^{i\pi y+y\log(x)}=e^{i\pi y}x^y$$

I am asking, since I worry that one can have more than one representation of $-1$:

$$-1=e^{i\pi+2\pi in}~~~,~~~n\in\mathbb{Z}$$

and $e^{i\pi y+2\pi iny}$ might have several distinct values for various $y$.

EDIT:

But I also feel like my worries are not really justified, since even for a positive number to a complex power we could write

$$x^y=(e^{\log(x)+2\pi i n})^y=e^{y\log(x)+2\pi i n y}=e^{2\pi i n y}x^y$$

Clearly, the extra $e^{2\pi i n y}$ appeared out of thin air and should not be there. After all, it is just $1^y=1$. So I'm inclined to think that $(-x)^y=e^{i\pi y}x^y$ is true as well. Any objections?

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I object.

$ 1^y $ is just $1$ only if $y$ is an integer. Otherwise, all the other possibilities cannot be ignored.

Check this question out as well: Are complex numbers subject to different rules of math?

You might also find my first blog article interesting and informative: https://www.dsprelated.com/showarticle/754.php

Ced

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  • $\begingroup$ I see, but how to reconcile this with the answer to the following question? math.stackexchange.com/questions/2616891/… $\endgroup$ – Kagaratsch May 4 '18 at 2:01
  • $\begingroup$ @Kagaratsch, As I stated in my answer to the other question I referenced. You have to have an understanding when principal values are being used and when all possible multiple answers are appropriate. The conditions stated in the reference you gave are for when you can assume the statements are true for the principle values, i.e. those rules won't lead to contradictions. $\endgroup$ – Cedron Dawg May 4 '18 at 2:11
  • $\begingroup$ OK, I see. So essentially, I am free to choose a branch of $1$ and stick to it as long as the operations I use do not lead to mixing of branches, as far as I understand. $\endgroup$ – Kagaratsch May 4 '18 at 2:15
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    $\begingroup$ @Kagaratsch, I wouldn't word it like that. It is more like when you use the quadratic equation that you check the possible answers for the one that is valid (or both). $\endgroup$ – Cedron Dawg May 4 '18 at 2:23

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