Question

If $e^{i\theta = \cos\theta + i\sin\theta$ then}
$\sum_{n=0}^{\infty} \dfrac{\cos(n\theta)}{2^n} \div \sum_{n=0}^{\infty} \dfrac{\cos(n\theta)}{5^n} =$

• A. $\dfrac{4 + 2\cos\theta}{5 - 4\cos\theta}$
• B. $\dfrac{4 - 2\cos\theta}{5 + 4\cos\theta}$
• C. $\dfrac{4 - 2\cos\theta}{5 - 4\cos\theta}$
• D. $\dfrac{4 + 2\cos\theta}{5 + 4\cos\theta}$

Hint:

Know and apply geometric series involving cosine functions for simplification in complex exponential problems.

Answer 1

The Correct Option is C

We use the identity: $\sum_{n=0}^{\infty} \cos(n\theta) r^n = \dfrac{1 - r\cos\theta}{1 - 2r\cos\theta + r^2}$
Apply it for $r = \dfrac{1}{2}$ and $r = \dfrac{1}{5}$ separately:
Numerator: $\dfrac{1 - \frac{1}{2}\cos\theta}{1 - \cos\theta + \frac{1}{4}} = \dfrac{2 - \cos\theta}{5 - 4\cos\theta}$
Denominator: $\dfrac{1 - \frac{1}{5}\cos\theta}{1 - \frac{2}{5}\cos\theta + \frac{1}{25}} = \dfrac{5 - \cos\theta}{25 - 10\cos\theta + 1}$
Taking the ratio and simplifying gives $\dfrac{4 - 2\cos\theta}{5 - 4\cos\theta}$