Trigonometric Summation

Consider the complex number

$z=\dfrac{cis\theta}{2}$ where $cis\theta=\cos\theta+i\sin\theta$ .

Note that $z^2=\dfrac{cis2\theta}{2^2}\ , z^3=\dfrac{cis3\theta}{2^3}$ and so on.

Consider the infinite series $S$ whose real part is the required sum.

$S=z+z^2+z^3+z^4+....$

This equation can be simplified to $S=\dfrac{z}{1-z}$

Let $Re(k)$ denote real part of the complex number $k$ .

Then, our required sum is

$Re\left(\dfrac{z}{z-1}\right)= Re\left(\dfrac{cis\theta} {2-cis\theta}\right)$ which on simplification yields

$Answer=\dfrac{2\cos\theta-1}{5-4\cos\theta}$

Thus the required value is $A+B+C+D= \boxed{12}$

• We know that $cis\theta$ = $\cos\theta + i\sin\theta$ . Therefore $cos\theta$ = $e^{i\theta}+ e^{-i\theta}/2$ . Thereby putting the value of $\cos\theta$ in the given expression and applying sum of infinite tems of an $GP$ with first term $a$ / $1-r$ we get the answer as $2\cos\theta - 1/ 5-4\cos\theta$ . Hence on comparing the answer to that of given in question we get the answer as $\boxed{12}$ .