Finding the real parts of this equation

**Matt1993** · November 6th 2012, 04:18 AM

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

**Prove It** · November 6th 2012, 05:00 AM

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,

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

Now simplify...

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