### Find all possible values of $i^{-2i}$ (NBHM 2013)

Find all possible values of $i^{-2i}$.

Solution:

Any complex number in the unit circle can be written as $e^{i\theta} = \cos(\theta) + i \sin(\theta)$ for some $\theta \in \Bbb{R}$. Since $i$ is in the imaginary axis we can take $i = e^{i \frac{\pi}{2}}$.

We have $e^{i\theta}$ is defined in terms of sin and cos and hence is $2\pi$ periodic. That is $e^{i(\theta+2n\pi)} = e^{i\theta}$ for $n \in \Bbb{Z}$. Therefore $i = e^{i \frac{\pi}{2}} = e^{(i(\frac{\pi}{2})+2n\pi))} = e^{i(\frac{(4n+1)\pi}{2})}$ and $i^{-2i} = e^{-2i \cdot i(\frac{(4n+1)\pi}{2})} = e^{2(\frac{(4n+1)\pi}{2})} = e^{(4n+1)\pi}$ where $n \in \Bbb{Z}$.

Hence, all possible values of $i^{-2i}$ are $\{\ldots, e^{-7\pi}, e^{-3\pi}, e^{\pi}, e^{5\pi}, \ldots \}$.

### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
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