What are the mathematical origins of i^i=e^(pi/2)?

[thharrimw]

could someone tell me where the i^i=e^(pi/2) came from??

 

[trambolin]

http://en.wikipedia.org/wiki/Imaginary_unit#Example

 

[n!kofeyn]

Ha. They're wrong. $i^i$=0.20787 and $e^{\pi/2}$=4.81047. They're missing a negative sign. Remember that taking the square root of something is the same thing as raising that something to the 1/2 power. Then $\sqrt{-1} = i = (-1)^{1/2}$. Also, 1/i=-i. Then
$$\begin{align*}<br /> e^{i\pi}+1 &= 0 \\<br /> e^{i\pi} &= -1 \\<br /> \left( e^{i\pi} \right)^{1/2} &= (-1)^{1/2} \\<br /> e^{i\pi/2} &= i \\<br /> \left( e^{i\pi/2} \right)^{1/i} &= i^{1/i} \\<br /> e^{\pi/2} &= i^{-i}<br /> \end{align*}$$

 

[GrizzlyBat]

I do not like the idea that $$i ^{\frac{1}{i}}$$ is $$i ^{-i}$$. I would of thought it would be $$i ^{i^{-1}}$$ which are not the same thing as far as I know.

 

[Cyosis]

It is true that most often they are not the same, however for i they are.

Proof:
$$\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$

 

[GrizzlyBat]

Oh cool, I didn't know that. That is pretty nice to know. And it makes sense when put like that.