# Math Help - Geometric series with complex numbers

1. ## 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 . I know it is really hard (at least for me ), 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)).

2. Originally Posted by juanma101285
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 . I know it is really hard (at least for me ), 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)).
Assuming that $1+z+z^2+\ldots+z^{n-1}=\dfrac{1-z^n}{1-z}$, represent the complex number $z=x+iy$ in exponential form $z=e^{i\theta}$ for some $\theta\in\mathbb{R}$ (recall that $e^{i\theta}=\text{cis}\,(\theta)=\cos\theta+i\sin\ theta$). Now substitute this into the geometric formula to get the first result.

To get from $\dfrac{1-e^{i n\theta}}{1-e^{i\theta}}=\dfrac{\sin\left(\frac{n\theta}{2}\ri ght)}{\sin\left(\frac{\theta}{2}\right)}e^{i(n-1)\theta/2}$, use $\sin\theta=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}$ and then recall DeMoivre's Theorem: $(\text{cis}\,(\theta))^n = \text{cis}\,(n\theta)$. This will involve a handful of manipulations to get to the result.

Finally, the last result is achieved by taking the imaginary part of each complex term (which is why everything is in terms of sine in the end).

I hope this kinda gives you an idea of how to do this. I hope this makes sense!!