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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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Badshah
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    Have a look at this, and in particular, look at the last subsection: http://en.wikipedia.org/wiki/Exponentiation#Complex_exponents_with_positive_real_bases. You will see that your answer is incorrect, and you will see why. – M Turgeon Oct 01 '12 at 18:32
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    Oke, I understand why its incorrect. Thanks – Badshah Oct 01 '12 at 18:39
  • Both answers are correct; there are infinitely many different values for $1^i$ that are all correct. – Carl Mummert Oct 02 '12 at 00:24
  • M Turgeon, thanks for that reference. The explanation there is a little beyond me but the example is great. I noticed that when you move the n back to the imaginary axis by factoring out the square the equation becomes true again, so is there a notational device to introduce that lets us use the laws of multiplication like we'd like to without having to mentally track when a trick may be used to force us to make the equation valid again without obstructing us - in the same way that complex numbers expanded our abilities in the first place? – codeshot Apr 26 '20 at 01:33

2 Answers2

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

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

Argon
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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 01 '12 at 18:55
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    @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 01 '12 at 19:05
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    @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 01 '12 at 19:29
  • thank you, this is very clear! – Badshah Oct 01 '12 at 20:07
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    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 01 '12 at 20:55
  • Yes, but usually when we take log of a real, we mean the real logarithm. 1 is real. So $\log(1) = 0$. But yes, technically the answer is $e^{-2\pi k}$, $k \in \mathbb{Z}$. – The_Sympathizer Oct 02 '12 at 00:27
  • It's not at all true that "when we take the log of a real, we mean the real logarithm" - when we are working in complex analysis, which we have to be in order to talk about $1^i$, the $\log$ function is the complex logarithm, not the real one. – Carl Mummert Oct 02 '12 at 00:29
  • I edited as I was receiving down votes for, I assume, not including other branches. – Argon Oct 02 '12 at 01:43
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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?

Berci
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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 above:http://en.wikipedia.org/wiki/Exponentiation#Complex_exponents_with_positive_real_bases – Badshah Oct 01 '12 at 20:07
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    @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 01 '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 02 '12 at 00:31
  • Why introduce k on the rhs and not the lhs? Isnt 1^i just equal to 1 over e^(2pi). Ie each branch of the former is equal to the corresponding branch of the latter? – codeshot Apr 24 '20 at 01:54
  • @codeshot I started from equation $1=e^{2k\pi i}$. – Berci Apr 24 '20 at 07:52
  • I suspect the starting equation is misleading, because also 1^ik = e^(2.pi.i). We should be able to describe this relationship in a way which doesn't treat one side of an equality more specially than the other. – codeshot Apr 24 '20 at 12:48