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Let $$S = e^{i\alpha} + \frac{e^{i3\alpha}}{3} + \frac{e^{i5\alpha}}{3^2} + \cdots$$

Find Im$(S)$ and show that it is equal to the sum

$$I = \sin(\alpha) + \frac{\sin(3\alpha)}{3} + \frac{\sin(5\alpha)}{3^2} + \cdots$$

So, I found that $S = \frac{3(3e^{i\alpha} - e^{-i\alpha})}{10 - 6\cos(2\alpha)}$ using the formula for geometric series.

I have a provided answer of $\frac{6\sin(\alpha)}{5 - 3\cos(2\alpha)}$ which I can see that I get if I just take the $\sin(\alpha)$ terms out of the $e$ terms in my numerator; however, why don't I have to change the $\cos(2\alpha)$ term in the denominator? Isn't this part of Re$(S)$? I was confused about his part and was trying to change this term before I looked at the answer.

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You could just say that the imaginary part of the sum is the sum of the imaginary parts (which is easily overkilled by Fubini), from which the equality easily follows. The value is calculated easily by geometric series formula, as you've stated. –  tomasz Jun 14 '12 at 2:20
    
Yah, I guess I am just wondering why there is a $\cos(\alpha)$ term in the imaginary part of a sum. Why did I get rid of the $\cos(\alpha)$ terms in the numerator, but leave the one in the denominator? –  stariz77 Jun 14 '12 at 2:22
    
If $5-3\cos(2\alpha)$ is real then it certainly appears in the denominator of the imaginary part as well. Just as $i/C$ is purely imaginary for any non-zero real constant $C$. –  M.B. Jun 14 '12 at 2:22
    
@M.B. ahh ok, so it's just like $i$ times a real number, making it imaginary. That real number being $\frac{1}{5 - 3\cos(2\alpha)}$ –  stariz77 Jun 14 '12 at 2:24

2 Answers 2

up vote 4 down vote accepted

$$ \begin{align} S &=e^{ia}\sum_{k=0}^\infty\left(\frac{e^{i2\alpha}}{3}\right)^k\\ &=\frac{3e^{i\alpha}}{3-e^{i2\alpha}}\cdot\frac{3-e^{-i2\alpha}}{3-e^{-i2\alpha}}\\ &=\frac{9e^{i\alpha}-3e^{-i\alpha}}{10-6\cos(2\alpha)}\\ &=\frac{6\cos(\alpha)+12i\sin(\alpha)}{10-6\cos(2\alpha)}\\ &=\color{red}{\frac{3\cos(\alpha)}{5-3\cos(2\alpha)}}+i\color{green}{\frac{6\sin(\alpha)}{5-3\cos(2\alpha)}} \end{align} $$ Therefore, $$ \mathrm{Im}(S)=\color{green}{\frac{6\sin(\alpha)}{5-3\cos(2\alpha)}} $$ Since the imaginary part of $\color{blue}{\dfrac{e^{i(2k+1)\alpha}}{3^k}}$ is $\color{orange}{\dfrac{\sin((2k+1)\alpha)}{3^k}}$ we get your second sum $$ \begin{align} \mathrm{Im}(S) &=\mathrm{Im}\left(\sum_{k=0}^\infty\color{blue}{\frac{e^{i(2k+1)\alpha}}{3^k}}\right)\\ &=\sum_{k=0}^\infty\color{orange}{\frac{\sin((2k+1)\alpha)}{3^k}} \end{align} $$

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$$ S = e^{i\alpha} + \frac{e^{i3\alpha}}{3} + \frac{e^{i5\alpha}}{3^2} + \cdots = e^{i\alpha}\left(1 + \frac{e^{2i\alpha}}{3} + \frac{(e^{2i\alpha})^2}{3^2}+ \cdots\right). $$ The thing inside the parentheses is a geometric series with common ratio $e^{2i\alpha}/3$. Therefore the expression above is equal to $$ \frac{e^{i\alpha}}{1-(e^{2i\alpha}/3)} = \frac{3e^{i\alpha}}{3-e^{2i\alpha}}. $$ This is $$ \frac{3e^{i\alpha}(3-e^{-2i\alpha})}{(3-e^{2i\alpha})(3-e^{-2i\alpha})} = \frac{9e^{i\alpha}- 3e^{-i\alpha}}{9 - 3e^{2i\alpha} - 3e^{-2i\alpha} + 1} = \frac{6\cos\alpha + 12i\sin\alpha}{10 - 6\cos(2\alpha)}. $$

That the imaginary part of this is equal to $i$ times that other sum follows from two facts: (1) the terms of the other sum are the imaginary parts of the terms of the first sum; and (2) the imaginary part of the sum equals the sum of the imaginary parts.

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I think you lost the $3$ somewhere. –  robjohn Jun 14 '12 at 4:16
    
@robjohn : I think I've fixed that. –  Michael Hardy Jun 16 '12 at 13:02
    
Indeed you have, but now $$ \begin{align} 9e^{i\alpha}-3e^{-i\alpha} &=9(\cos(\alpha)+i\sin(\alpha))-3(\cos(\alpha)-i\sin(\alpha))\\ &=6\cos(\alpha)\color{red}{+}12i\sin(\alpha) \end{align} $$ –  robjohn Jun 16 '12 at 14:10
    
Haste makes waste . . . . . –  Michael Hardy Jun 16 '12 at 17:00

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