Finding the $i$th power of $i$

Question

I have been trying to find the $i$th power of $i$ as a mental exercise, I have tried two approaches

For the first, using properties of exponents

$$i^i=i^{\sqrt{-1}}=i^{-1^{\frac{1}{2}}} \implies i^i=i^{-\frac{1}{2}}=\frac{1}{i^2}=-1$$

However this seems a bit, for a lack of a better word "bodgy".

The next method I used was rewriting into exponential form

$$z=i^i + 0 \implies \arg(z)=\tan^-{1}\left(\frac{1^i}{0}\right)=\frac{\pi}{2}$$
$$ \therefore i^i=e^{\frac{\pi}{2}i}$$

I am quite sure that none of these answers are correct, what are the flaws in my method and how should I go about actually computing this ?

See also this post (not quite the same question but obviously related) — Ben Grossmann, Apr 18 '21 at 11:04
For both of your approaches, I do not understand the $\implies$ step. It would be helpful if you tried to explain your thought process there — Ben Grossmann, Apr 18 '21 at 11:05
You've conflated $a^{b^c}$ with $a^{bc}$, which doesn't even work with $a=b=2,,c=3$. — J.G., Apr 18 '21 at 11:09
@BenGrossmann For the first step I applied the exponent property $(x^a)^b=x^{ab}$ and for the next I used the property that $tan(z)=\frac{y}{x}$ and since the coefficient of the imaginary part is $1$ and 1 raised to any power is 1 I got the argument as $\frac{\pi}{2}$ — Filthyscrub, Apr 18 '21 at 11:11
@Filthyscrub Thanks for clarifying. For the first step, as J.G. notes, $$ i^{(-1)^{1/2}} \neq [i^{-1}]^{1/2}. $$ For the second, it's not clear where $1^i$ came from — Ben Grossmann, Apr 18 '21 at 11:13

score 3 · Answer 1 · answered Apr 18 '21 at 11:08

The typical approach to the problem is this: $$ i = e^{\pi i /2} \implies i^i = [e^{\pi i/2}]^i = e^{\pi i^2 /2} = e^{-\pi/2}. $$ However, as I note in my post here, this actually leads to several (infinitely many) possible values for $i^i$. Interestingly, each of these possibilities is a positive real number.

score 0 · Answer 2 · answered Apr 18 '21 at 11:12

0

Write

$$i^i = \left(e^{\log i}\right)^i = e^{i \log i}$$

and use the fact that

$$ \log w = \mathrm{ln}|w| + i \,\mathrm{arg}\,w$$

Note that $\mathrm{arg}\,w$ is a multivalued function.

answered Apr 18 '21 at 11:12

Andy Walls

3,461

Finding the $i$th power of $i$

2 Answers2