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I am trying to solve this problem, and show that it does not necessarily equal $i^{-1}$. My method so far is $$(i)^{i} \\ =e^{(i)\ln{(i)} } i = \cos(\theta)+i\sin(\theta) \Rightarrow \theta = \frac{\pi}{2}\pm2n\pi, n \in \mathbb{Z} \\ \Rightarrow i = e^{i\frac{\pi}{2}\pm2in\pi} \\ \Rightarrow \ln{(i)} = i\frac{\pi}{2}\pm2in\pi \\ \Rightarrow e^{(i)\ln{(i)} } = e^{(i)(i\frac{\pi}{2}\pm2in\pi)}\\ =e^{(-\frac{\pi}{2}\pm2n\pi)}$$

However when I raise this to the power of $i$ once more, the answer ends up becoming $i^{-1}$ once more. What am I missing here? If someone can please show me where I'm going wrong that would be fantastic!

John Page
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1 Answers1

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This question is a duplicate of a previous one, however the answers does not really explain in detail what is going on with respect to OPs calculations so I'll add a CW here to try address this.

First let's define properly what we mean by $(i^i)^i$. The standard way to define this is by taking $z^a \equiv e^{a\text{Log}_n(z)}$ where $\text{Log}_n(z)$ is the $n$'th branch of the logarithm defined by

$$\text{Log}_n(z) = \log|z| + i\arg(z) + 2\pi i n\tag{1}$$

We then have $$(i^i)^i \equiv e^{i\,\text{Log}_n(e^{i\,\text{Log}_n(i)})}$$

Now $i\,\text{Log}_n(z) = -\frac{\pi}{2} - 2\pi n$ is real however the logarithm $\text{Log}_n$ of the real number $e^{i\,\text{Log}_n(z)}$ does not have to be. This is where the mistake is in your calculation. Using $(1)$ above we see that

$$\text{Log}_n(e^{i\,\text{Log}_n(i)}) = -\frac{\pi}{2} - 2\pi n + 2\pi in$$

so

$$(i^i)^i \equiv e^{-\frac{\pi i}{2} - 2\pi i n - 2\pi n} = \frac{e^{-2\pi n}}{i}$$

Only for the principal branch $n=0$ is this equal to $\frac{1}{i}$.

Winther
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