About numbers to the powers of imaginary numbers

Question

I once saw a problem which asks for the real and imaginary parts $(i^i)^i$. The problem also implied that there are several solutions to this. However I am very confused, $(i^i)$ is a real number so isn't its imaginary part just that real number raised to the power of $i$? Please help thanks!

Answer 1

$$i^i=\left(e^{i\frac{\pi}{2}}\right)^i=e^{-\frac{\pi}{2}}$$ therefore $$\left(i^i\right)^i=\left(e^{-\frac{\pi}{2}}\right)^i=e^{-i\frac{\pi}{2}}=\cos\left(\frac{\pi}{2}\right)-i\sin\left(\frac{\pi}{2}\right)=-i$$ and so $\Re\left(\left(i^i\right)^i\right)=0$ and $\Im\left(\left(i^i\right)^i\right)=-1$

Answer 2

$$\begin{align}(i^i)^i & = i^{i\times i} \\ & = i^{-1}\\ & =e^{-i(2k+\frac{1}{2})\pi}, (k\in\mathbb Z)\\ & =-i\end{align}$$

If on the other hand you meant $i^{i^i}$, we have: $$\begin{align} i^{i^i} & = i^{e^{(i(2m+\frac{1}{2})\pi)i}} \\ & = i^{e^{-(2m+\frac{1}{2})\pi}}\\ & = e^{i[(2n+\frac{1}{2})\pi e^{-(2m+\frac{1}{2})\pi}]}, (m,n\in\mathbb Z)\end{align}$$

for the top answer the real part is obviously $0$ and the imaginary part $-1$; for the bottom answer:

$${\frak Re} (i^{i^i})=\cos((2n+\frac{1}{2})\pi e^{-(2m+\frac{1}{2})\pi}), (m,n\in\mathbb Z)\\{\frak Im} (i^{i^i})=\sin((2n+\frac{1}{2})\pi e^{-(2m+\frac{1}{2})\pi}), (m,n\in\mathbb Z)$$

Answer 3

@Zima 's answer clearly answers this question, with the appropriate principal arguments. However, in this text (not answer) I am going talk more about your statement of "several solutions".

Recall that polar plane is closed and angles following $\theta + 2\pi$ repeat the same output for $\theta$; $\sin{\theta}$, $\cos{\theta}$. Ie. $\sin{\dfrac{\pi}{2}} = \sin{\left(\dfrac{\pi}{2} + 2\pi\right)} = \sin{\dfrac{5\pi}{2}}$ and $\cos{\dfrac{\pi}{2}} = \cos{\left(\dfrac{\pi}{2} + 2\pi\right)} = \cos{\dfrac{5\pi}{2}}$.

This property exists within the complex world as well, however, we note it as "branching" and equivalent to your multivalued solutions, since the polar form $\displaystyle e^{i\theta} = \cos{\theta} + i\sin{\theta}$. In practice, this would mean;

$$\left(e^{i\pi/2}\right)^i = i^i = \left(e^{i5\pi/2}\right)^i \\\iff i^i = e^{-2n\pi - \pi/2}, \quad n \in \mathbb{Z}$$

However, note that $\theta = \dfrac{\pi}{2}$ is still the principal argument for $i$. Any other angle that has equivalent output to $\theta = \dfrac{\pi}{2}$, would classified under $\underbrace{\text{general arguments}}_{\theta = 2n\pi + \pi/2}$. See Complex Analysis.

I assume this is what your problem also implied that there are several solutions. However, I am still confused, since this is not required to answer the question. Your problem asks for the real and imaginary parts (i^i)^i, which I assume would be in rectangular form; $0 + (-1)i$ only. Did the question explicitly ask you to state all possible arguments somewhere else?