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The number $\operatorname{cis} 75 + \operatorname{cis} 83 + \operatorname{cis} 91 +\dots+ \operatorname{cis} 147$ is expressed in the form $r\operatorname{cis}(\theta)$, where $0\leq \theta< 360$. Find $\theta$ in degrees

I'm having major trouble with this problem.

Michael Hardy
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Aditya
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  • Use the fact that $\left(\mbox{cis}(1)\right)^\theta=\mbox{cis}(\theta)$, which is true by De Moivre. Then, your series is a geometric one. – Akiva Weinberger Aug 24 '14 at 03:10
  • @John, See http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – lab bhattacharjee Aug 24 '14 at 05:11

2 Answers2

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Hint: you want to evaluate $$e^{i(75\pi/180)}+e^{i(83\pi/180)}+e^{i(91\pi/180)}+\cdots +e^{i(147\pi/180)}.$$

This is a geometric series with common ratio $r=e^{i\frac{8\pi}{180}}.$

Now use the formula for the sum to $n$ terms of a geometric series: $$S_n=\frac{1-r^n}{1-r}.$$

Also, don't forget to convert back to degrees using

$$\boxed{\theta^\circ =\theta^{\ \rm{rad}}\times \frac{180}{\pi}}.$$

beep-boop
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  • I love your answer. However, the OP tagged this as Precalculus, and it is probable that he does not understand $\operatorname{cis}(\theta)$ as $e^{i\theta}$. Is there a solution that uses just DeMoivre's theorem perhaps? – Rory Daulton Aug 23 '14 at 23:38
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Rearrange the 10 terms as: $$ (\operatorname{cis} 75 + \operatorname{cis} 147) + (\operatorname{cis} 83 + \operatorname{cis} 139) + \cdots + (\operatorname{cis} 107 + \operatorname{cis} 115)$$ By the parallelogram rule, each pair of terms here is the diagonal of a rhombus which goes in the direction halfway between the two angles -- that is, in this case always parallel to the direction $\frac{75+147}{2} = 111$. All that remains to see is whether the actual direction is $111$ or its opposite.

However, if the angles are in degrees, then we easily see that all of the original terms have positive imaginary part -- and therefore so must their sum, so the answer is $111^\circ$.

hmakholm left over Monica
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