proof

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March 1st 2012, 01:03 PM
Mhmh96

proof

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March 1st 2012, 01:19 PM
Plato

Re: proof

To prove this you need to know the principal value of a complex exponential.
$z^w=\exp(w\cdot\text{Log}(z))$

Recall that $\text{Log}(i)=\ln(1)+\frac{i\pi}{2}.$
March 1st 2012, 01:47 PM
Mhmh96

Re: proof

Ok,but how can i start?
March 1st 2012, 02:11 PM
Plato

Re: proof

Quote:

Originally Posted by Mhmh96

Ok,but how can i start?

What do you mean by that?

What does $i\text{Log}(i)=~?$ .

Then what is $\exp(i\text{Log}(i))~?$
March 1st 2012, 02:19 PM
Mhmh96

Re: proof

Yes, i have no background about how to prove it so i want to know the proof in details ,if that ok.
March 1st 2012, 02:28 PM
Plato

Re: proof

Quote:

Originally Posted by Mhmh96

i have no background about how to prove it so i want to know the proof in details ,if that ok.

Well that is not going to happen here.

Because if you have have no background then you would not begin to understand the proof.
March 1st 2012, 02:34 PM
Mhmh96

Re: proof

I guess that is right,thanks anyway.
March 2nd 2012, 07:34 AM
Mhmh96

Re: 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?
March 4th 2012, 07:01 AM
Soroban

Re: proof

Hello, Mhmh96!

Quote:

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

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

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

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

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

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

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

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

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