# Exp((-i*pi)/4) = ?

1. Sep 10, 2009

### Bacat

Not really homework, but part of a homework problem I am working on.

I know that $$e^{i\pi}+1=0$$ (Euler's Identity)

And also that $$e^{i \pi} = e^{-i \pi}$$

But I'm having trouble understanding $$e^{\frac{-i \pi}{4}}$$

In the complex plane this is a clockwise rotation around the origin of $$\frac{\pi}{2}$$ radians. But I think it should reduce to some real constant which I am having trouble finding.

In Mathematica, I get two different answers...

$$N[e^{\frac{-i \pi}{4}}] = 0.707107 - 0.707107 i$$

This implies that $$e^{\frac{-i \pi}{4}} = \frac{1}{\sqrt{2}}(1-i)$$ which seems wrong to me.

Simplify$$[e^{\frac{-i \pi}{4}}] = -(-1)^{\frac{3}{4}}$$

But this reduces to 1, which I believe is probably the correct answer.

Is the first result just spurious rounding?

Can I just write the following identity as true?

$$e^{\frac{-i \pi}{4}} = 1$$

2. Sep 10, 2009

### Dick

e^(-i*pi/4) isn't 1 by a long shot. e^(i*x) is cos(x)+i*sin(x). e^(-i*pi/4)=cos(-pi/4)+i*sin(-pi/4). Both of your Mathematica answers are correct. In fact they are the same (after rounding). And (-1)^(3/4) is NOT 1. Or -1 either.

3. Sep 11, 2009

### snipez90

Eh the first answer looks correct to me, since .707 is around sqrt(2)/2. You do know the Euler's formula, which is where those identities are usually derived from right?

4. Sep 11, 2009

### Staff: Mentor

No, the rotation is by $$\frac{\pi}{4}$$ radians. Was the 2 in the denominator a typo?
Why would you think that? Just because $$e^{-i \pi}$$ is a real constant, doesn't mean that the other one is also a real constant.
These are two representations of the same complex number. The first is an approximation and the second is exact.
I am not familiar with Mathematica, so I don't know the difference between the N command and the Simplify command.
No it doesn't, and 1 is not the correct answer. Let's look at -(-1)3/4 a little more closely. Before the final sign change, you have (-1) to the 3/4 power. That is the same as -1 cubed (still -1), which we take the 4th root of. This is not a real number, since there is no real number that when squared, and then squared again, yields a negative number. The final step is to change the sign of this (nonreal) number, which still doesn't give us 1, as you claimed.
Absolutely not.

5. Sep 11, 2009

### Bacat

Thanks Mark44 and all. This makes a lot more sense to me now.