2. ## 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 $\text{Log}(i)=\ln(1)+\frac{i\pi}{2}.$

3. ## Re: proof

Ok,but how can i start?

4. ## Re: proof

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 $\exp(i\text{Log}(i))~?$

5. ## Re: proof

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

6. ## Re: proof

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.

7. ## Re: proof

I guess that is right,thanks anyway.

8. ## 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?

9. ## Re: proof

Hello, Mhmh96!

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

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

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

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

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

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

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

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

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