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At one point, I found an equation that works with complex logarithms, but I lost the book that contains the equation. If I feed this to Wolfram|Alpha, it states that $1^{-i}$ is equal to 1. Why is $1^{-i} = 1$?

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  • $\begingroup$ because $a^z = \exp(z\log a) $ for any real number $a$? $\endgroup$
    – mookid
    Nov 11, 2014 at 18:17
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    $\begingroup$ Note that we have to be careful with our definition of complex exponentiation. I could claim with equal consistency that $$ 1^{-i} = (e^{2 \pi i})^{-i} = e^{2 \pi} $$ $\endgroup$ Nov 11, 2014 at 18:22
  • $\begingroup$ @Omnomnomnom Isn't it somewhat standard to take the $(-\pi,\pi]$ branch of the complex logarithm? $\endgroup$ Nov 11, 2014 at 18:23
  • $\begingroup$ @user2345215 yes; however, it's occasionally useful to switch branches though (for example, taking $[0,2 \pi)$), and that choice often depends on the context. $\endgroup$ Nov 11, 2014 at 18:25
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    $\begingroup$ Well $1^z$ is a multi valued function, right? $\endgroup$ Nov 11, 2014 at 18:25

2 Answers 2

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Since $a^b = e^{b \log a}$, we have

$$1^{-i} = e^{-i \log 1} = e^{-i \cdot 2k\pi i} = e^{2k\pi}$$

Note that in the complex numbers $\log^\mathbb C z = \log^\mathbb R |z| + (2k \pi + \arg z)i$, so there are infinite choices for its value.

One of the values that $1^{-i}$ takes is $1$, but it is not always $1$. If we set $k = 0$, the resulting number it's called principal value (http://en.wikipedia.org/wiki/Principal_value), and that is what wolfram reports.

With $k=0$ you get $e^0 = 1$

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  • $\begingroup$ This leaves out the key fact that there are many different choice for $\log(1)$ in the complex numbers. $\endgroup$ Nov 11, 2014 at 18:27
  • $\begingroup$ @CarlMummert You are correct. It should be fixed now $\endgroup$
    – Ant
    Nov 11, 2014 at 18:30
  • $\begingroup$ There is a typo in the revision. $e^{-i\log i} = e^{-i\cdot 2k\pi i} = e^{2k\pi}$. This will given different values depending on the value of $k$; it's not always $1$. But it is $1$ when $k =0$; this corresponds to the principal value. $\endgroup$ Nov 11, 2014 at 18:32
  • $\begingroup$ @CarlMummert Thanks! I was superficial :) It is okay now? $\endgroup$
    – Ant
    Nov 11, 2014 at 18:35
  • $\begingroup$ It seems OK, except the last non-equality is sometimes an equality, depending on what $k$ is. $\endgroup$ Nov 11, 2014 at 18:41
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Given any $a,b\in \mathbb{C}$ we define $a^b=e^{b\ \log\ (a)}$. Thus in this case $1^{-i}$ = $e^{-i\log (1)}$ = $1$.

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  • $\begingroup$ This leaves out the key fact that there are many different choice for $\log(1)$ in the complex numbers. $\endgroup$ Nov 11, 2014 at 18:27
  • $\begingroup$ @CarlMummert that's true; i assumed that we are talking about the principal value, as the question which has been asked is that why is $1^{-i}=1$, not that what are the possible values for $1^{-i}$. $\endgroup$
    – wanderer
    Nov 11, 2014 at 18:31
  • $\begingroup$ That is the question, but the answer is "in general, it's not". It is certainly true that the principal value of $1^{-i}$ is $1$, of course. $\endgroup$ Nov 11, 2014 at 18:33
  • $\begingroup$ @CarlMummert right!! $\endgroup$
    – wanderer
    Nov 11, 2014 at 18:33

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