5

I watched this video by Matt Parker recently: https://www.youtube.com/watch?v=9tlHQOKMHGA

He calculates $i^i$ and his answer is around ~1/5 well more precise $e^\frac{-\pi}{2}$ which uses

$e^{i\theta} = \cos(\theta)+i\sin{\theta}$

where $e^{i\frac{\pi}{2}} = i$

so that $i^i = e^{\ln{i^i}} = e^{i\ln{i}} = e^{ii\frac{\pi}{2}} = e^{-\frac{\pi}{2}}$ which is around 1/5

whatever. If you use

$e^{i\theta} = \cos(\theta)+i\sin{\theta}$

you can also argue that

$e^{i\frac{5\pi}{2}} = i$

and so

$i^i = e^{-\frac{5\pi}{2}}$

which is obviously different to $e^{-\frac{\pi}{2}}$.

Where is the error?

WolframAlpha btw agrees with Matt ;)

http://www.wolframalpha.com/input/?i=i%5Ei

Asaf Karagila
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Wikunia
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4 Answers4

8

There is no error.

There are actually infinite values for $i^i$ due to the periodicity.

Enrico M.
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  • Hmm yeah that kind of makes sense. It's just weird in my head :D – Wikunia Dec 15 '17 at 17:26
  • @Wikunia Haha it kinda is, because how can an expression assume infinite different values? But it's that beautiful! $i^i$ is REAL!! – Enrico M. Dec 15 '17 at 17:27
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    Haha and yeah it is REAL in for every of it's infinite values. That makes it kind of complex ;) – Wikunia Dec 15 '17 at 17:30
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    This is not really a correct answer. In the same vein you could also say that $\sqrt 4 = -2$ is true – but it's not, because by convention only the positive sign is considered. Likewise, by convention we only consider the principal branch of the logarithm, and that implies that $i^i = e^{-\pi/2}$, nothing else. – leftaroundabout Dec 15 '17 at 19:02
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    @leftaroundabout It's not "by convention". It's just because you use the most standard branch cut. But in any case $i^i$ has not a single defined value. Remember that we also have

    $$i = e^{5 i \pi/2}$$

    Which gives a different value.

     – Enrico M. Dec 15 '17 at 19:09
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    $e^{i\pi/2}$ and $e^{5i\pi/2}$ are the same value, by the axioms of the complex numbers. Again, this is completely analogous to $2^2$ and $(-2)^2$ – leftaroundabout Dec 15 '17 at 19:11
5

$i^i$ is by definition $e^{i \log i}$ where $\log z = \log |z| + i \arg z$ is a branch of log.

If you choose your branch of log to be $-\pi < \arg z < \pi$, then $i^i = e^{-\frac{\pi}{2}}$.

If you choose your branch of log to be $\pi < \arg z < 3 \pi$, then $i^i = e^{-\left(\frac{\pi}{2} + 2 \pi\right)}$

So different branch of log gives you different value of $i^i$.

The branch of log with $-\pi < \arg z < \pi$ is called the principal branch, which is what people usually assumed if the branch of log is not specified. In this branch, the value of $i^i$ is called the principal value of $i^i$, denoted by $\operatorname{PV} i^i$

If you want to know more this kind of stuff, you can take a course in complex number / complex analysis.

Alex Vong
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3

The secret is on how you define logarithm for complex numbers, if we have a number in the unit circle $x$, we choose its logarithm by convention to be the $0\leq\theta<2\pi$ for which $e^{2\pi i \theta}=x$; otherwise log(1) could be equal to $2k\pi i$ for any integer $k$.

Marcelo Campos
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1

There is no error. If $a,b\in\mathbb C$ and $a\neq0$, then $a^b$ stands for any numbers of the form $\exp\bigl(b\log(a)\bigr)$, where $\log a$ can be any logarihtm of $a$.

José Carlos Santos
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