Summation of complex numbers

I-Think · #1 $Old$ November 1st, 2009, 09:37 PM

Given that $w_n=3^{-n}cos2n\theta$ for $n=1,2,3$ ... use De Moivre's Theorem to show that

$1+w_1+w_2+w_3+...+w_{N-1}=\frac{9-3cos2\theta+3^{-N+1}cos2(N-1)\theta-3^{N+2}cos2N\theta}{10-6cos2\theta}$

I am really hopelessly lost on this question.
Help is greatly appreciated.
Thanks in advance.

red_dog · #2 $Old$ November 2nd, 2009, 08:53 AM

Let $S_1=1+\frac{1}{3}\cos 2\theta+\frac{1}{3^2}\cos 4\theta+\ldots+\frac{1}{3^n}\cos 2n\theta$

and $S_2=\frac{1}{3}\sin 2\theta+\frac{1}{3^2}\sin 4\theta+\ldots+\frac{1}{3^n}\sin 2n\theta$

Then $S_1+iS_2=1+\frac{1}{3}(\cos 2\theta+i\sin 2\theta)+\frac{1}{3^2}(\cos 4\theta+i\sin 4\theta)+\ldots+\frac{1}{3^n}(\cos 2n\theta+i\sin 2n\theta)$

Let $z=\frac{1}{3}(\cos 2\theta+i\sin 2\theta)$

Then $S_1+iS_2=1+z+z^2+\ldots+z^n$

Now use the sum of a geometric progression and put the right hand member in the form $A+Bi$ . Then $S_1=A, \ S_2=B$