How do I sum a geometric series with complex numbers?

[HmBe]

Homework Statement

Homework Equations

a(1-r^[n+1])/(1-r)

The Attempt at a Solution

So I wrote it as e^(-iNz) [1 + e^(iz) + e^(2iz) + ... + e^(2iNz)]

Let r = e^(iz), a=e^(-iNz)

a [1 + r + r^2 + ... + r^(2N)]

From here I'm not sure what to do. I tried letting n=2N, and then letting the bracket be equal to (1-r)^-1 but I couldn't get much out of that. Tried using the formula, can't get anything. I just get loads of e's everywhere, and no way to get rid of the cos if I put them in trig form. Really stuck.

 

[awkward]

Once you have applied the formula for the sum of a geometric series, you will have a fraction. The trick is to multiply the numerator and denominator of the fraction by an appropriate factor. It's hard to say more without giving the whole thing away, but you can probably work backwards from the answer if you can't figure out the factor otherwise.

 

[vela]

Hint: Use $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$. Can you tweak the numerator and denominator to look like that using awkward's suggestion?

 

[HmBe]

Thanks a lot for the help, totally sorted now. Had forgotten that identity, was all I needed, cheers you two.