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For the geometric progression in complex variable z : $$1 + z + z^2 +...+ z^{(n-1)} = (1-z^n)/(1-z)$$ Using the polar representation $$z = \cos(\theta) + i\sin(\theta), $$ Need show that: $$(a) 1 + 2\cos\theta +2\cos2\theta +...+ 2\cos(n-1)\theta = \frac{\sin(n -1/2)\theta}{\sin\theta/2}$$ $$(b)\sin\theta + \sin2\theta +...+ \sin(n-1)\theta = \frac{\cos\theta/2 - \cos(n-1/2)\theta}{2\sin\theta/2} $$

I have found that direct substitution of the polar form for z in the geometric series is too cumbersome. Instead, either some sort of trick is needed; or an alternative technique is needed. If simple type of mathematical induction is tried, then it would tantamount to working from reverse.

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  • $\begingroup$ a) is easy to prove using induction...I didn't try b.. $\endgroup$ – Isham Sep 20 '17 at 8:16
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HINT:

Put $z=e^{i\theta}$

Now we know How to prove Euler's formula: $e^{it}=\cos t +i\sin t$?

$$\dfrac{1-z^n}{1-z}=\dfrac{1-e^{in\theta}}{1-e^{i\theta}}=e^{i(n-1)\theta/2}\cdot\dfrac{e^{in\theta/2}-e^{-in\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}=\left(\cos\dfrac{(n-1)\theta}2+i\sin\dfrac{(n-1)\theta}2\right)\dfrac{2i\sin\dfrac{n\theta}2}{2i\sin\dfrac{\theta}2}$$

Now equate the real & the imaginary parts.

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HINT: use De Moivre formula, you have $z^k = \cos (k\theta) + i\sin(k\theta)$.

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