Geometric series with complex numbers

Hi, I have the following exercise and I have no clue how to approach it...I would really appreciate it if anyone could give me a hand since I have never done this type of exercises (Doh). I know it is really hard (at least for me (Angry)), but any help will be highly appreciated. Thanks!

The formula for the sum of a geometric series holds for complex numbers too. Thus z∈C (complex numbers) and n∈N (natural numbers) it is true that

1+z+z^2+...+z^(n-1)=(1/z^n)/(1-z).

Explain briefly how we can see that for θ∈R (real numbers) we have

1+e^(iθ)+2^(2iθ)+...+e^[(n-1)iθ]=(1-e^(inθ))/(1-e^(iθ)).

Write down the formula for sinθ in terms of e^(iθ) and e^(-iθ). Show that

(1-e^(inθ))/(1-e^(iθ))=(sin((nθ)/2))/(sin(θ/2)) * e^(i(n-1)θ/2)

and hence show that

sin(θ) + sin(2θ) + sin(3θ) + ... sin((n-1)θ)=(sin((nθ)/2)*sin((n-1)θ/2))/(sin(θ/2)).