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What is the value of $1^i$?

I was thinking, what would 1^i be? Then I did: $e^{i\pi}=-1\rightarrow e^{i\pi}\cdot e^{i\pi}=e^{2i\pi}=-1\cdot -1=1$ Now raise to the power i: $1^i=(e^{2i\pi})^i=e^{2i^2\pi}=e^{-2\pi}=\frac{1}{e^{2\pi}}$ Is this correct?

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marked as duplicate by Carl Mummert, Pedro Tamaroff, Rahul, Argon, Sasha Oct 2 '12 at 4:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have a look at this, and in particular, look at the last subsection:…. You will see that your answer is incorrect, and you will see why. – M Turgeon Oct 1 '12 at 18:32
Oke, I understand why its incorrect. Thanks – Badshah Oct 1 '12 at 18:39
Both answers are correct; there are infinitely many different values for $1^i$ that are all correct. – Carl Mummert Oct 2 '12 at 0:24
up vote 11 down vote accepted

$$1^i = e^{i\log 1}=e^{i(\log |1|+i\arg 1)}=e^{i(i 2\pi n)}=e^{-2\pi n}$$

Where the principal branch of the logarithm is given by $n=0$.

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This isn't the "only" answer though, since you still have the freedom to choose which branch of the logarithm to use. – Christopher A. Wong Oct 1 '12 at 18:55
@ChristopherA.Wong I think for real numbers, the exponential is defined using the principal branch. I don't think there's a freedom in choice in this case. Otherwise wouldn't the OP's solution also be valid? – EuYu Oct 1 '12 at 19:05
@EuYu, there's no reason why that should be the case. For exponentiating positive numbers to real powers, it doesn't even matter what branch of the logarithm to pick. And yes, it is my opinion that the OP's solution is valid when specifying the second branch of the logarithm. – Christopher A. Wong Oct 1 '12 at 19:29
thank you, this is very clear! – Badshah Oct 1 '12 at 20:07
While this is certainly the most sensible answer in most cases, and agrees with convention, it is not hard to come up with a scenario in which this answer is problematic: Draw a curve from the origin into the fourth quadrant, crossing into the first quadrant between $1$ and $2$ on the real axis, then going upwards to infinity. One can define a continuous branch of the logarithm on the complement of this curve. Its value must deviate from the convention either at $z=1$ or at $z=2$. – Harald Hanche-Olsen Oct 1 '12 at 20:55

The power function $(x,y)\mapsto x^y$ is indeed defined via the exponential $x\mapsto e^x$, as $x^y:=e^{\log x\cdot y}$, where $\log$ is the inverse of $\exp$. But, $\exp$ is not injective: it is periodic by $2\pi i$, hence the $\log$ is not unique, only up to $+k2\pi i$ for some $k\in\mathbb Z$. So, $1^i$ has infinite many values: $$1^i=e^{(2k\pi i)i} = e^{-2k\pi} $$ Similarly for $i^i$. Can you find all of its values?

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i dont think you can do exponentiation like that: (a^b)^c is'nt always a^(bc) for complex numbers, see… – Badshah Oct 1 '12 at 20:07
@Badshah But this answer does not use the formula you quote. Indeed, any sensible definition of $x^y$ has to be of the form $e^{y\log x}$ for some choice of branch for the logarithm. Just look a bit further down on the wikipedia page you quoted. – Harald Hanche-Olsen Oct 1 '12 at 20:23
@Badshah: in the complex case, $a^b$ is defined as $e^{b\log a}$. The issue is that $\log 1$ is not in general $0$ when we talk about the complex logarithm. Silently assuming that $\log 1 = 0$ can lead to many errors in the complex setting. – Carl Mummert Oct 2 '12 at 0:31

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