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-   -   Where is $i^i$ ? (http://mymathforum.com/complex-analysis/340659-where-i-i.html)

 agentredlum May 25th, 2017 02:44 PM

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 2-dimensional unit circle. Seems like we lost a dimension.

What is going on here?

 v8archie May 25th, 2017 03:15 PM

Well $i$ is detached from $\frac{a}{b}$, so it's not surprising that there's a discontinuity there is it?

 Country Boy May 25th, 2017 03:21 PM

. since so . Then and , a real number.

(That is the "principal value". You can get others by adding to the .)

 agentredlum May 27th, 2017 05:43 AM

Here is a follow-up 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 $0$?

??

:)

 v8archie May 29th, 2017 04:52 PM

$$i^z = (e^{i\frac\pi2})^{a+bi} = e^{-b\frac\pi2}e^{ai\frac\pi2}$$

That's the whole complex plane.

 agentredlum May 30th, 2017 03:33 AM

Except the origin of course. If you are right , we just found a simple equation for the entire complex plane sans the origin.

 Country Boy September 29th, 2017 05:32 AM

That should have been
Quote:
 Originally Posted by Country Boy (Post 571098) $\displaystyle i^i= e^{ln(i^i)}= e^{i ln(i)}$. since $\displaystyle i= e^{i\frac{\pi}{2}}$ so $\displaystyle ln(i)= i\frac{\pi}{2}$. Then $\displaystyle i ln(i)= -\frac{\pi}{2}$ and $\displaystyle i^i= e^{-\frac{\pi}{2}}$, a real number. (That is the "principal value". You can get others by adding $\displaystyle 2\pi i$ to the $\displaystyle i\frac{\pi}{2}$.)

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