Where is $i^i$ ? For $ \ \ \mathbb{a} \ \ , \ \ \mathbb{ b} \ \ € \ \ \mathbb{ Z} \ \ , \mathbb{ b} \ne 0 $ We know where $ \ \ i^ { \frac{a}{b} } \ \ $ is. It is on the unit circle centered at the origin in the complex plane. It is not hard to imagine that for $ \ \ \mathbb{ a} \ \ , \mathbb{ b} \ \ € \ \ \mathbb{ R} \ \ , \ \ \mathbb{ b} \ne 0 \ \ , \ \ i^ { \frac{a}{b} } \ \ $ fills up all the points on the unit circle centered at the origin in the complex plane. Now , $ i^i \ \ $ is not on this unit circle. It is on the real number line detached from the 2dimensional unit circle. Seems like we lost a dimension. What is going on here? 
Well $i$ is detached from $\frac{a}{b}$, so it's not surprising that there's a discontinuity there is it? 
. since so . Then and , a real number. (That is the "principal value". You can get others by adding to the .) 
Here is a followup question: Consider $f(x) = i^x$, $x \ \in\ \mathbb{C}$ The base is $i$ and the domain of the exponent $x$ is all complex numbers. What is the range? Or to put it differently, we know for $\ x \ \in\ \mathbb{R} \ $ the output is on the unit circle centered at the origin in the complex plane. We also know the output can fall on the real axis, for example $\ f(i) = i^i$. Where else can the output wind up? Can the output be any point in the complex plane? Can $ \ \ i^x = a + bi \ \ $ for all $ \ \ a$, $b \ \in \ \mathbb{R} $ except possibly when $a$, $b$ are both simultaneously $0$? ?? :) 
$$i^z = (e^{i\frac\pi2})^{a+bi} = e^{b\frac\pi2}e^{ai\frac\pi2}$$ That's the whole complex plane. 
Except the origin of course. If you are right , we just found a simple equation for the entire complex plane sans the origin. 
That should have been Quote:

All times are GMT 8. The time now is 11:14 PM. 
Copyright © 2019 My Math Forum. All rights reserved.