# Eulers idenity

1. Jul 14, 2009

### thharrimw

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

Last edited by a moderator: Sep 25, 2014
2. Jul 14, 2009

### trambolin

3. Jul 15, 2009

### 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*} e^{i\pi}+1 &= 0 \\ e^{i\pi} &= -1 \\ \left( e^{i\pi} \right)^{1/2} &= (-1)^{1/2} \\ e^{i\pi/2} &= i \\ \left( e^{i\pi/2} \right)^{1/i} &= i^{1/i} \\ e^{\pi/2} &= i^{-i} \end{align*}

4. Jul 15, 2009

### 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.

5. Jul 15, 2009

### Cyosis

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

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

6. Jul 15, 2009

### GrizzlyBat

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