Finding the real parts of this equation

Finding the real parts to both sides of this equation

I am really stuck trying to find the real parts to both sides of this equation.

1 + exp(theta(i)) + exp(2theta(i)) + .... + exp((n-1)theta(i)) = (exp(ntheta(i)) - 1) / (exp(theta(i)) -1)

I think that the real part of the left hand side is

1 + cos(theta) + cos(2theta) + ..... + cos((n-1)theta)

i have no idea what the real part is for the right.

Someone please help

Re: Finding the real parts of this equation

Quote:

Originally Posted by

**Matt1993** Finding the real parts to both sides of this equation

I am really stuck trying to find the real parts to both sides of this equation.

1 + exp(theta(i)) + exp(2theta(i)) + .... + exp((n-1)theta(i)) = (exp(ntheta(i)) - 1) / (exp(theta(i)) -1)

I think that the real part of the left hand side is

1 + cos(theta) + cos(2theta) + ..... + cos((n-1)theta)

i have no idea what the real part is for the right.

Someone please help

I agree with the answer to your LHS. As for the RHS,

$\displaystyle \displaystyle \begin{align*} \frac{e^{n\theta i} - 1}{e^{\theta i} - 1} &= \frac{\cos{(n\theta)} + i\sin{(n\theta)} - 1}{\cos{(\theta)} + i\sin{(\theta)} - 1} \\ &= \frac{\left[ \cos{(n\theta)} - 1 + i\sin{(n\theta )} \right] \left[\cos{(\theta)} - 1 - i\sin{(\theta)}\right]}{\left[ \cos{(\theta )} - 1 + i\sin{(\theta )} \right] \left[ \cos{(\theta )} - 1 - i\sin{(\theta )} \right]} \end{align*}$

Now simplify...