http://img825.imageshack.us/img825/4720/capturejnt.jpg

Uploaded with ImageShack.us

Printable View

- Mar 1st 2012, 01:03 PMMhmh96proof
- Mar 1st 2012, 01:19 PMPlatoRe: proof
To prove this you need to know the principal value of a complex exponential.

$\displaystyle z^w=\exp(w\cdot\text{Log}(z))$

Recall that $\displaystyle \text{Log}(i)=\ln(1)+\frac{i\pi}{2}.$ - Mar 1st 2012, 01:47 PMMhmh96Re: proof
Ok,but how can i start?

- Mar 1st 2012, 02:11 PMPlatoRe: proof
- Mar 1st 2012, 02:19 PMMhmh96Re: proof
Yes, i have no background about how to prove it so i want to know the proof in details ,if that ok.

- Mar 1st 2012, 02:28 PMPlatoRe: proof
- Mar 1st 2012, 02:34 PMMhmh96Re: proof
I guess that is right,thanks anyway.

- Mar 2nd 2012, 07:34 AMMhmh96Re: proof
I think i found the proof ,but i still have one more problem

when we raise i to the power i ,the result should be real number,right? - Mar 4th 2012, 07:01 AMSorobanRe: proof
Hello, Mhmh96!

Quote:

$\displaystyle \text{Prove: }\:i^i \:=\:\dfrac{1}{\sqrt{e^{\pi}}}$

$\displaystyle \text{Let }\,z \:=\:i^i$

$\displaystyle \text{Take logs: }\:\ln(z) \:=\:\ln(i^i) \quad\Rightarrow\quad \ln(z)\:=\:i \ln(i) \;\;{\bf[1]}$

$\displaystyle \text{Given: }\:e^{i\theta} \:=\:\cos\theta + i\sin\theta$

$\displaystyle \text{Let }\theta = \tfrac{\pi}{2}\!:\;\;e^{i\frac{\pi}{2}} \;=\;\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2} \quad\Rightarrow\quad e^{i\frac{\pi}{2}} \:=\:i $

$\displaystyle \text{Take logs: }\:\ln\left(e^{i\frac{\pi}{2}}\right) \:=\:\ln(i) \quad\Rightarrow\quad i\tfrac{\pi}{2}\ln(e) \:=\:\ln(i) $

. . $\displaystyle \text{Hence: }\:\ln(i) \:=\:i\tfrac{\pi}{2}$

$\displaystyle \text{Substitute into }{\bf[1]}\!:\;\;\ln(z) \:=\:i\left(i\tfrac{\pi}{2}\right) \quad\Rightarrow\quad \ln(z)\:=\: -\tfrac{\pi}{2}$

$\displaystyle \text{Therefore: }\:z \;=\;e^{-\frac{\pi}{2}} \;=\;\dfrac{1}{e^{\frac{\pi}{2}}} \;=\; \dfrac{1}{(e^{\pi})^{\frac{1}{2}}} \;=\;\dfrac{1}{\sqrt{e^{\pi}}} $