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I am given a complex number $z=i^{2i}$ so what would be $|z|$? Firstly I took $\log$

$$ \log(z)=2i\log i\\ $$ thus $$ \log(z)=2i\log(e^{-π/2}) = -i\pi; $$ then inverting the log $$ z=e^{ -i\pi} $$ implying $$ |z| =1.$$

Is this correct?

Joe
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Tesla
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4 Answers4

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All of the values of $i^{2i}$ are real. The possible values are $$i^{2i} = e^{2i\cdot\log i}=e^{2i\cdot(\frac{\pi}{2}i+2k\pi i)} =e^{-\pi -4k\pi}=e^{-(4k+1)\pi} $$ which are all positive, so they are all possible values of $|z|$.

Note that none of these is $1$.

MPW
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Remember:

  • By definition, $i^{2i} = \exp(2i \cdot \log(i))$.

  • There are infinitely many different possible values of $\log(i)$. Because of the way $\exp(z)$ works, we have $e^{\pi i/2} = i$ and $e^{2\pi i n} = 1$ for all $n \in \mathbb{Z}$. So the values of $\log(i)$ are of the form $\pi i/2 + 2\pi i n = i\pi(1/2 +2n)$ for $n \in \mathbb{Z}$.

The combination of these facts (that complex powers are defined relative to a complex logarithm, and that the complex logarithm in general has more than one possible value) cause complex exponents to behave very differently from real ones (see here for another example: $1^i$ is not always $1$).

In the case at hand, $i^{2i}$, we can choose any integer $n$ in the logarithm of $i$, obtaining $$ i^{2i} = \exp(2i \cdot \pi i (1/2 + 2n)) = \exp(-2\pi(1/+2n)) = \exp((-1-4n)\pi) $$

The exponent there is completely real, so the value of the exponential function will match the value of the real exponential function. You can see that there are infinitely many different possible values for $|i^{2i}|$, one for each value of $\log(i)$ (that is, one for each value of $n$).

Community
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Carl Mummert
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$i=e^{\frac12\pi i}$ so that $$i^{2i}=(e^{\frac12\pi i})^{2i}=e^{\frac12\pi i.2i}=e^{-\pi}$$

edit

This answer is not complete. See the comments of Carl. It only computes the so-called principle value. This by picking the usual logarithm. Be aware of the confusion that can arise here. I was a victim of that myself.

drhab
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  • This is only part of the story; see the answer by MPW – Carl Mummert Feb 20 '15 at 14:44
  • @CarlMummert Yes, I have seen that, and realize that it is indeed not the whole story. I am still wondering though where things go wrong in my answer. – drhab Feb 20 '15 at 14:46
  • The issue is that $i$ is also $e^{5\pi i/2}$, $e^{9\pi i /2}$, etc. This answer finds the principal value by picking the "usual" logarithm. – Carl Mummert Feb 20 '15 at 14:48
  • @CarlMummert Thank you. Of course you mean $i$ here (not $1$). I will delete this answer within some minutes. – drhab Feb 20 '15 at 14:49
  • Yes, thanks. But it would not be bad to just leave a note that this computes the principal value - this is a common point of confusion (e.g. this very question) – Carl Mummert Feb 20 '15 at 14:50
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    @CarlMummert I solved it by means of this edit. – drhab Feb 20 '15 at 14:55
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Your first line is not correct, the property $\ln x^a=a\ln x$ is only valid for real numbers. However it is true that $e^{\ln x^a}=e^{a\ln x}$. So, in the end, your answer is correct.

Tim Raczkowski
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    I'm pretty sure that $\ln x^a=a\ln x$ is also valid for complex numbers. – barak manos Feb 20 '15 at 14:24
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    Actually it's not. Consider the branch of $\ln$ for which $-\pi\le\text{arg }z<\pi$. $\ln (e^{3\pi i/4})^2=\ln(-i)=-\frac\pi 2$, but $2\ln e^{3\pi i/4}=\frac{3\pi i}2.$ – Tim Raczkowski Feb 20 '15 at 14:29
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    @barakmanos No it does not. You have to take into account the branch of the log , since it is a multivalued function. But what does hold is:

    $$\log z_1^{z_2} =z_2 \log z_1 + 2\pi i k , ;; k \in \mathbb{Z}$$

     – Tolaso Feb 20 '15 at 14:33