If $z=i^{2i}$ then $|z| =?$ 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?
 A: 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 $1$.
A: 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 $n$).
A: $i=e^{\frac12\pi i}$ 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.
A: 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.
