# Homework Help: Fourier series complex numbers

1. Jun 17, 2012

### robertjford80

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I don't understand this equation. 2pi, 4pi, 6pi only = 0 when there is a sine function before it, so I don't see how the evens = 0. I don't see why the e vanishes. I also can't get the i's to vanish since one of them is in exponential form and the other is not. i * 2^i does not get rid of the i in my book. really lost here.

2. Jun 17, 2012

### Staff: Mentor

No. None of the numbers 2$\pi$, 4$\pi$, 6$\pi$, etc. is equal to zero. It's true that sin($2\pi$) = 0, but the argument is not equal to zero.

$e^{-i\pi n}$ happens to be equal to 1 for any even integer n. This is because of Euler's formula, eix = cos(x) + isin(x). Replace x by any even multiple of $\pi$ to see this.

3. Jun 17, 2012

### robertjford80

I'm watching this lecture,

so if I still don't understand after watching it, I'll get back to you.

Last edited by a moderator: Sep 25, 2014
4. Jun 17, 2012

### robertjford80

I can't figure out how to combine cos(x) + isin(x) to get a real number. For instance 2 + i is still imaginary.

5. Jun 17, 2012

### Infinitum

Think what happens when sin(x) is zero....(equivalent to x=n∏)

In simpler terms, what happens to (a+ib) if b is zero

6. Jun 17, 2012

### robertjford80

if b is 0 in a + ib then we just get a line along the x axis, y = 0. I don't see that helps me with my problem.

7. Jun 17, 2012

### Infinitum

No, you don't. You get 'a', a real number, which is constant. Not a line.

By Euler's formula, $e^{i\theta} = cos\theta + i\cdot sin\theta$ which is the same as $z = a+ib$

See now?

8. Jun 17, 2012

### robertjford80

I don't see how euler's formula gets the equation I listed in the OP

9. Jun 17, 2012

### Infinitum

Well....the equation you have listed in the OP contains $$e^{-in\pi}$$ and Euler's formula gives an alternate expression for $$e^{i\theta}$$ Is there any resemblance?

10. Jun 17, 2012

### robertjford80

no resemblance. even if there was it wouldn't tell me why 1/πin, n odd, nor would it tell me why (e^-inπ -1)/-2πin results in 1/πin, n odd or 0 if n is even

11. Jun 17, 2012

### Infinitum

But but...its all laid out to resemble

Another hint : put θ = -n∏, in Euler's equation....

Also, try to forget about the (1/-2∏in) term for the time being.

12. Jun 17, 2012

### robertjford80

never mind, i'm tired of your hints. i give up.

13. Jun 17, 2012

### Infinitum

Good idea. You should take a break and try again when you are fresh, sometime. :tongue2:

Its very, very simple

14. Jun 17, 2012

### Staff: Mentor

No, 2 + i is complex, which means it has a real part (2) and an imaginary part (1, the coefficient of i). 2 + 0i is purely real. 2i = 0 + 2i is purely imaginary, and its imaginary part is 2.