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The roots of $z^7 = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$ are $\text{cis } \theta_1, \text{cis } \theta_2, \dots, \text{cis } \theta_7,$ where $ 0^\circ \le \theta_k < 360^\circ $for all $ 1 \le k \le 7$. Find $\theta_1 + \theta_2 + \dots + \theta_7$. Give your answer in degrees.

In exponential form this is $z^7 = e^ \left(5 \pi i/4 \right)$. How should I simplify? Thanks

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What is $\text{cis}$? Do you by chance mean $\cos$? –  AlexR Jan 24 '14 at 16:26
    
@AlexR cis is short for $cos(\theta)+isin(\theta)$ –  Mr.Fry Jan 24 '14 at 16:27
    
@AlexR: "cis" is a commonly used abbreviation for "cosine plus $i$ times the sine." In other words, $$\operatorname{cis}\theta=\cos\theta+i\sin\theta=e^{i\theta}.$$ –  Cameron Buie Jan 24 '14 at 16:27

3 Answers 3

Well, by DeMoivre's Theorem, we have $$(\operatorname{cis}\theta_k)^7=\operatorname{cis}(7\theta_k)$$ for $k=1,...,7.$ We need for $$(\operatorname{cis}\theta_k)^7=\operatorname{cis} 225^\circ,$$ as you've already determined, and so we need $$\operatorname{cis}(7\theta_k)=\operatorname{cis} 225^\circ\\\frac{\operatorname{cis}(7\theta_k)}{\operatorname{cis} 225^\circ}=1\\\operatorname{cis}(7\theta_k-225^\circ)=1$$ for $k=1,...,7.$ Can you take it from there?

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If I am understanding correctly...for $k=1$, we take $$\operatorname{cis}(7\theta_1)=\operatorname{cis} 225^\circ$$ and then somehow determine $\theta_1$, and repeat for $k=2,3,4,5,6,7$? –  Mathy Person Jul 18 at 21:59
    
No, that's getting further from the answer. Rather, think about the solutions set for the equation $$\operatorname{cis}\theta=1.$ This will let you get rid of the $\operatorname{cis},$ and find all solutions to the original equation. Then you can find the seven solutions in the desired interval. –  Cameron Buie Jul 18 at 22:10
    
I'm not sure exactly what you mean. Can you elaborate? –  Mathy Person Jul 18 at 22:18
    
Well, what are the solutions to $$\operatorname{cis}\theta=1,$$ or, more usefully put, $$\cos\theta+i\sin\theta=1+0i$$ –  Cameron Buie Jul 18 at 22:24
    
$\cos \theta = 1$ and $\sin \theta = 0$, so $\theta$ = 360 degrees/$2\pi$. –  Mathy Person Jul 18 at 22:30

The roots are $e^{2k\pi i/7}$ ($k=0,\ldots,6$). Write them as $r^k$ where $r = e^{2\pi i/7}$. So the sum is (using the simple formula for summing a finite geometric series)

$$\sum_{k=0}^6 r^k = \frac{1-r^7}{1-r} = \frac{1 - e^{2\pi i}}{1-e^{2\pi i/7}} = 0$$.

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This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not summing up the roots, but rather the arguments of said roots lying in $[0^\circ,360^\circ).$ –  Cameron Buie Jan 24 '14 at 17:15

$\def\cis{\operatorname{cis}}$Let's consider a more general problem. We have $a=r\cis\alpha$ and we write its $n$th roots as \begin{gather} \sqrt[n]{r}\cis\frac{\beta}{n}\\ \sqrt[n]{r}\cis\left(\frac{\beta}{n}+\frac{2\pi}{n}\right)\\ \sqrt[n]{r}\cis\left(\frac{\beta}{n}+2\frac{2\pi}{n}\right)\\ \dots\\ \sqrt[n]{r}\cis\left(\frac{\beta}{n}+(n-1)\frac{2\pi}{n}\right) \end{gather} Can you sum up all those angles? Use $360$ instead of $2\pi$ if you so prefer or are forced to.

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