Expansion of geometric progression of complex variable in polar coordinates

Question

The below problem is taken from the text by: James Victor Uspensky, titled: Theory of Equations; sec. 1.14.

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.

Answer 1

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.

Answer 2

Let $z = e^{i \theta}$,

From $$\sum_{j=0}^{n-1}z^j = \frac{1-z^n}{1-z},$$

We have $$\sum_{j=0}^{n-1}e^{ij\theta} = \frac{1-e^{i \theta n}}{1-e^{i \theta}}= \left( \cos \frac{(n-1) \theta }2 + i \sin \frac{(n-1) \theta}{2}\right) \frac{\sin \frac{n\theta}2}{\sin \frac{\theta}2} \tag{1}$$

If we take the real part of $(1)$, we get

$$\sum_{j=0}^{n-1} \cos (j \theta) = \cos \frac{(n-1) \theta}2 \cdot \frac{\sin \frac{n \theta}{2}}{\sin \frac{\theta}2}$$

\begin{align}2\sum_{j=0}^{n-1} \cos (j \theta) -1= 2\cos \frac{(n-1) \theta}2 \cdot \frac{\sin \frac{n \theta}{2}}{\sin \frac{\theta}2}-1 = \frac{2\cos \frac{(n-1)\theta}2\sin \frac{n \theta}2 - \sin \frac{\theta}2}{\sin \frac{\theta}2}\tag{2}\end{align}

Notice that we have $\sin (n-\frac12)\theta + \sin \frac{\theta}2 = 2 \cos \frac{(n-1) \theta}2 \sin \frac{n \theta}2 $ by the sum and product formula, hence $ 2 \cos \frac{(n-1) \theta}2 \sin \frac{n \theta}2 - \sin \frac{\theta}2 = \sin (n - \frac12) \theta$. Substitute this in $(2)$ and we obtain

$$2\sum_{j=0}^{n-1} \cos (j \theta) -1=\frac{\sin (n-\frac12) \theta}{\sin \frac{\theta}2}$$ which is $(a)$.

To prove $(b)$, we take the imaginary part of $(1)$,

$$\sum_{j=1}^{n-1} \sin j \theta = \sin \frac{(n-1) \theta}2 \cdot \frac{\sin \frac{n\theta}2}{\sin \frac{\theta}2}$$

Now from the sum and product formula, we have $\cos \frac{\theta}2 - \cos (n-\frac12) \theta= 2 \sin \frac{n \theta}2 \sin \frac{(n-1)\theta}2$, hence we have

$$\sum_{j=1}^{n-1} \sin j \theta = \frac{\cos \frac{\theta}2 - \cos (n-\frac12) \theta}{2\sin \frac{\theta}2}$$

which is $(b)$.

Answer 3

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