# proof

• Mar 1st 2012, 01:03 PM
Mhmh96
proof
• Mar 1st 2012, 01:19 PM
Plato
Re: 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 PM
Mhmh96
Re: proof
Ok,but how can i start?
• Mar 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 $\displaystyle i\text{Log}(i)=~?$.

Then what is $\displaystyle \exp(i\text{Log}(i))~?$
• Mar 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.
• Mar 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.
• Mar 1st 2012, 02:34 PM
Mhmh96
Re: proof
I guess that is right,thanks anyway.
• Mar 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?
• Mar 4th 2012, 07:01 AM
Soroban
Re: proof
Hello, Mhmh96!

Quote:

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

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

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

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

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

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

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

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

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