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

by Bacat
Tags: complex, exponential, numerical, rotation, rounding
 P: 151 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. The other answer given is: 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? $$e^{\frac{-i \pi}{4}} = 1$$
 HW Helper Sci Advisor Thanks P: 24,460 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.
 P: 1,106 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?
Mentor
P: 19,762

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

 Quote by 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.
No, the rotation is by $$\frac{\pi}{4}$$ radians. Was the 2 in the denominator a typo?
 Quote by Bacat But I think it should reduce to some real constant which I am having trouble finding.
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.
 Quote by Bacat 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.
These are two representations of the same complex number. The first is an approximation and the second is exact.
 Quote by Bacat The other answer given is: Simplify$$[e^{\frac{-i \pi}{4}}] = -(-1)^{\frac{3}{4}}$$
I am not familiar with Mathematica, so I don't know the difference between the N command and the Simplify command.
 Quote by Bacat But this reduces to 1, which I believe is probably the correct answer.
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.
 Quote by Bacat Is the first result just spurious rounding? Can I just write the following identity as true? $$e^{\frac{-i \pi}{4}} = 1$$
Absolutely not.
 P: 151 Thanks Mark44 and all. This makes a lot more sense to me now.