Why does this work for $ i^{2i} $?

Question

I'm finding the principal value of $$ i^{2i} $$

And I know it's solved like this:

$$ (e^{ i\pi /2})^{2i} $$

$$ e^{ i^{2} \pi} $$

$$ e^{- \pi} $$

I understand the process but I don't understand for example where does the $ i $ in $ 2i $ go?

Is this some kind of a property of Euler's number? if so please explain to me.

Answer 1

$$\bigl(e^{i\pi /2}\bigr)^{2i} = e^{(i\pi /2) \cdot 2i} = e^{i^2\pi}.$$

This is just an application of the exponent laws. Don't overthink it!

Answer 2

The $i$ in $2i$ was combined with the $i$ inside the parentheses. Hence, you got $$i\cdot i = i^2$$ which is due to exponent laws. More applied to your case: $$(e^{ i\pi /2})^{2i}=e^{i \cdot i \cdot \pi}=e^{i^2\cdot \pi}.$$

Answer 3

Also remember that $i = e^{\pi i/2+2\pi i k} =e^{\pi i(\frac12+2k)} $ for any integer $k$, since $e^{2 \pi i k} = 1$. $k=0$ gives the usual principal value.

Therefore, $i^{2i} =(e^{\pi i(\frac12+2k)})^{2i} =e^{2\pi i^2(\frac12+2k)} =e^{-2\pi (\frac12+2k)} =e^{-\pi (1+4k)} $. So, thanks to the joy of infinite valued complex logarithms, your expression has an infinite number of distinct values.

Note that, for large negative $k$, the value is quite large.