I'm having trouble with part $ii.$ of the following question:

$i.$ Express the following in terms of N and z: $$\sum^N_{n=1}2^{-n}z^n$$

Expanding with geometric series: $$\sum^N_{n=1}2^{-n}z^n \equiv \sum^N_{n=1}\bigg(\frac{z}2\bigg)^n = \bigg(\frac{z}2\bigg) + \bigg(\frac{z}2\bigg)^2 + \bigg(\frac{z}2\bigg)^3 + ... + \bigg(\frac{z}2\bigg)^N$$ $$= \frac{\bigg(\frac{z}2\bigg)\bigg(1-\bigg(\frac{z}2\bigg)^N\bigg)}{1 - \bigg(\frac{z}2\bigg)} = \frac{z\bigg(1-\bigg(\frac{z}2\bigg)^N\bigg)}{2-z}$$

$ii.$ Deduce that: $$\sum^{10}_{n=1}2^{-n}\sin(\frac1{10}n\pi) = \frac{1025\sin(\frac1{10}\pi)}{2560-2048\cos(\frac1{10}\pi)}$$

Looking instead at the imaginary part of the following series: $$\sum^{10}_{n=1}2^{-n}e^{i\frac{n\pi}{10}} = \frac12e^{i\frac{\pi}{10}} + \frac14e^{i\frac{2\pi}{10}} + \frac18e^{i\frac{3\pi}{10}} + ... + \frac1{2^{10}}e^{i\pi}$$ $$=\frac{\bigg(e^{i\frac\pi{10}}\bigg)\bigg(1-\bigg(\frac{e^{i\frac\pi{10}}}2\bigg)^{10}\bigg)}{2-e^{i\frac\pi{10}}} = \frac{\bigg(e^{i\frac\pi{10}}\bigg)\bigg(1-\frac{e^{i\pi}}{2^{10}}\bigg)}{2-e^{i\frac\pi{10}}}$$

Applying $-e^{i\pi} \equiv 1$ and simplifying, then applying $e^{i\theta}\equiv r(\cos\theta+i\sin\theta)$: $$\implies \sum^{10}_{n=1}2^{-n}e^{i\frac{n\pi}{10}} = \frac{\bigg(e^{i\frac\pi{10}}\bigg)\bigg(2^{10}+1\bigg)}{2^{10}(2-e^{i\frac\pi{10}})} = \frac{1025(\cos\frac\pi{10}+i\sin\frac\pi{10})}{2048-1024(\cos\frac\pi{10}+i\sin\frac\pi{10})}$$

My understanding is that, from the final equation, looking at it's imaginary part we would have to take the $i$ term from the numerator and the real term from the denominator, like so: $$= \frac{1025(\sin\frac\pi{10})}{2048-1024(\cos\frac\pi{10})}$$

But how do I satisfy the denominator to get the required $(2560 - 2048\cos\frac\pi{10})$?

Edit: On second thought, adding $512-1024\cos\frac\pi{10}$ to the denominator does not balance the equation, which means that this is not the correct sum.


1 Answer 1



$$\frac{\cos t+i\sin t}{2-\cos t-i\sin t}$$

$$=\frac{(\cos t+i\sin t)(2-\cos t+i\sin t)}{(2-\cos t)^2+\sin^2 t}$$

$$=\frac{(2\cos t-1+2i\sin t)}{5-4\cos t}$$

  • $\begingroup$ Okay, simply multiplying $\frac{1025}{1024}$ to your simplification I got the required answer. Thanks a bunch. $\endgroup$
    – Ozzy
    Apr 25, 2013 at 16:52
  • 1
    $\begingroup$ @Ozzy, my pleasure. $\endgroup$ Apr 25, 2013 at 16:54

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