A Complex Infinite Trig Sum 2

Let $P = 1 + \frac {1}{x} \cos \theta + \frac {1}{x^2} \cos 2 \theta + \frac {1}{x^3} \cos 3 \theta...$

Let $Q = \frac {1}{x} \sin \theta + \frac {1}{x^2} \sin 2 \theta + \frac {1}{x^3} \sin 3 \theta$

(Note that these two series converge)

$P + iQ = 1 + \frac {1}{x} (\cos \theta + i\sin\theta) + \frac {1}{x^2} (\cos 2 \theta + i\sin2\theta) + \frac {1}{x^3} (\cos 3 \theta + i\sin3\theta)...$

Where $i = \sqrt{-1}$

$P + iQ = 1 + \frac {1}{x} e^{i \theta} + \frac {1}{x^2} e^{i 2\theta} + \frac {1}{x^3} e^{i 3\theta}$

This is a geometric sequence with a common ratio of $\frac {1}{x} e^{i \theta}$ .

So the converging series is equal to $\frac {1}{ 1 - \frac {1}{x}e^{i \theta}} = \frac {x}{ x -e^{i \theta}} = \frac {x}{( x -\cos \theta) - i\sin\theta}$

Rationalising the denominator we get $\frac {x^2 - x\cos\theta + ix\sin\theta}{x^2 + 1 - 2x\cos\theta}$

Taking the real part of the equation, $P = \frac {x^2 - x\cos\theta}{x^2 + 1 - 2x\cos\theta}$ .

Substituting $\theta = 60$ : $P = \frac {2x^2 - x}{2x^2 - 2x + 2}$ .

So $\frac {2x^2 - x}{2x^2 - 2x + 2} - 1$ $=$ $\frac {2(x+1)^2 - (x+1)}{2(x+1)^2 - 2(x+1) + 2} - 1$

Solving for $x$ we find that $x^2 -3x -1 = 0$

$x = \frac {3+\sqrt{13}}{2}$

Therefore $a + b + c = \boxed{18}$

Using the fact $cosα=Re(e^{iα})$ (where $i=√(-1)$ ) and summing the series on each side (they are infinite G.P. series each) we get $x^2-3x-1=0$ . Solving this we get the positive value of $x$ equal to $\dfrac{3+√13}{2}$ . So $a=3, b=13, c=2$ . Hence $a+b+c=3+13+2=18$