Complex variables, De moivres formula

[MaxManus]

Homework Statement

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)}

The Attempt at a Solution

First I I am 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)}

Homework Statement

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

 

[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.

 

[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?

 

[hunt_mat]

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

 

[ehild]

Choose r=1.

ehild

 

[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.

 

[hunt_mat]

What we're saying is that:
<br /> \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})<br />
The denominator should now be a real number and you are free to choose the imaginary part at your leisure.

 

[hunt_mat]

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

 

[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)