# Homework Help: Complex variables, De moivres formula

1. Sep 22, 2010

### MaxManus

1. The problem statement, all variables and given/known data
z is a complex number different from 1 and n >= 1 is an integer

$$1 + z + z^2+ ..... + z^n = \frac{z^{n+1} - 1}{z-1}$$

show that:
$$\sin(\theta) + \sin(2 \theta)+ .... \sin(n \theta) = \frac{ \sin(n \theta/2) \sin((n+1) \theta / 2)}{\sin(\theta / 2)}$$

3. The attempt at a solution

First I Im both sides and De Moivres formula

$$Im( 1 + z + z^2+ ..... + z^n) = r \sin( \theta) + r^2 \sin(2 \theta) + ... r^n \sin(n \theta)$$

$$Im( \frac{z^{n+1} - 1}{z-1} ) = \frac{r^n \sin((n+1) \theta)}{\sin(\theta)}$$
1. The problem statement, all variables and given/known data

Can anyone give me a hint from here or tell me if I am on the wrong track

2. Sep 22, 2010

### hunt_mat

Almost, use $$z=re^{i\theta}$$ in your complex geometric progression and use De Morevre's theorem and take the imaginary part to obtain the solution.

3. Sep 23, 2010

### MaxManus

Thanks, I get the left side, but how do I simplefy the right side?
Left side

$$Im( 1 + r e^{ \theta} + ... + r^n e^{ \theta})$$
$$= sin( \theta) + ..... sin( n \theta)$$

right side

$$Im( \frac{ r^{n+1} e^{(n+1) \theta} -1}{r e^{ \theta} -1})$$
How do I take the imaginary part from this?

4. Sep 23, 2010

### hunt_mat

The RHS side should be (from your LaTeX)
$$\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}$$
Multiply by the complex conjugate of the denominator and take the imaginary part.

5. Sep 23, 2010

### ehild

Choose r=1.

ehild

6. Sep 23, 2010

### MaxManus

Thanks, but not sure if I understand
$$Im( \frac{ r^{n+1} e^{ i (n+1) \theta} -1}{r e^{i \theta} -1})$$

The complex conjugate $$r e ^{i (- \theta)}$$

$$Im( \frac{ r^{n+2} e^{ i n \theta} - r e^{i (- \theta)}{r^2 -r e^{i (- \theta)}})$$

Something wrong with my Latex, but what I have done i multiplied the denominator and the numerator with $$r e ^{i (-\theta}$$ and I still have a complet denominator.

Last edited: Sep 23, 2010
7. Sep 23, 2010

### hunt_mat

What we're saying is that:
$$\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}=\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\cdot (e^{-i\theta}-1)/(e^{-i\theta}-1})$$
The denominator should now be a real number and you are free to choose the imaginary part at your leisure.

Last edited: Sep 23, 2010
8. Sep 23, 2010

### hunt_mat

I can't get the Latex to work but the fraction should read as:
$$\frac{e^{-i\theta}-1}}{e^{-i\theta}-1}}$$

9. Sep 24, 2010

### MaxManus

Thanks^for all the help, I get it now.
----------------------------------------------
$$(e^{i \theta} -1)(e^{- i \theta} -1)$$
$$= (1 - e^{i \theta} -e^{- i \theta} +1)$$
$$= 1 - cos( \theta) - i sin( \theta) - cos( - \theta) - i sin( - \theta) +1$$
$$= 2 - 2cos(\theta)$$

---------------------------------------
$$cos( \theta) = cos( - \theta)$$
$$sin(- \theta) = - sin( \theta)$$