complex numbers: $i^i $ vs $\pi $: which is larger?

Question

I am kind of novice to complex numbers, may one pls check if the following makes sense: I am to compare $i^i$ vs $\pi$ ?

What is bizarre to me is that $i^i$ is complex while pi is real, so how come that two quantities are comparable directly .. curious where is my mistake?

$$i^i = k\text{, assume } k \in \mathbb{R}$$ $$\implies i\ln i = \ln k$$ $$i = 0 + 1i = \cos \theta + i\sin \theta$$ $$\implies r=1, \theta = \dfrac{\pi}{2} \text{ in } z = e^{i\dfrac{\pi}{2}}\text{ for polair form of } z=i$$ $$i^i = i\cdot \ln e^{i\dfrac{\pi}{2}}= i^2\dfrac{\pi}{2} = -\dfrac{\pi}{2} = \ln k$$ $$\implies e^{-\dfrac{\pi}{2}} = k = 0.20788$$ $$k < \pi$$

Answer 1

Don't be fooled by the presence of $i$, you wouldn't say $i^2 \not\in \mathbb{R}$, right?

What is bizarre to me is that $i^i$ is complex while $\pi$ is real, (...)

As you found with your own calculations, $i^i$ is in fact real but to get that value, you (implicitly) picked the principle branch of the (multi-valued) complex logarithm.

Picking that branch leads to the value $e^{-\pi/2}$, which is obviously smaller than $\pi$, but this isn't necessarily the case when picking another branch.

In general, you have: $$i^i = e^{- \frac{\pi}{2}+2\pi k} \quad (k \in \mathbb{Z})$$ where the principle value corresponds to taking $k=0$.

See also: Mathworld: Complex Exponentiation

Answer 2

$i^i$ is not complex. It is real number. You can write it as $(e^{i\frac \pi 2})^i$. Which becomes $e^{-\pi/2}$. As $e^x$ is increasing function, and $-\pi/2 \lt 0$ So $i^i \lt e^0 =1 \lt \pi$