Function of A Complex Variable

[darkchild]

Homework Statement

If$$z=e^{2\pi i/5}$$, then $$1+z+z^{2}+z^{3}+5z^{4}+4z^{5}+4z^{6}+4z^{7}+4z^{8}+5z^{9}=$$

(A) 0

(B) $$4e^{3\pi i/5}$$

(C) $$5e^{4\pi i/5}$$

(D) $$-4e^{2\pi i/5}$$

(E)$$-5e^{3\pi i/5}$$

Homework Equations

$$e^{2\pi i}=\cos(2\pi)+isin(2\pi)=1$$

The Attempt at a Solution

I plugged $$z=1$$ into the equation and calculated 30. None of the answer choices is equal to 30. I'm thinking that maybe I have to do something with the fifth roots of unity, but I'm not sure what.

 

[Dick]

You might want to use that the sum of the five fifth roots of unity is zero.

 

[spamiam]

Studying for the Math GRE, eh?

Anyway, I was just going to add that the fact that Dick mentioned comes from the factorization of $z^5 - 1$. Letting $\zeta = e^{{2 \pi i}/{5}}$ (I want z for my variable), which is a primitive 5th root of unity, we have

$$z^5 - 1 = (z - 1)(z- \zeta)(z - \zeta^2)(z - \zeta^3)(z - \zeta^4)$$

Calculating the coefficient for z⁴ yields the identity.