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Argand Diagram and \(\sum_{k-0}^{n-1} \cos \left(\frac {2k+1}{2n+1}\pi \right) = \frac 12\)

In a few of my solutions for trigonometry problems, I mentioned that Argand diagram shows that:

  • \(\displaystyle \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} \cos{\frac{(2k-1)\pi}{n}} = \frac{1}{2} \) and
  • \(\displaystyle \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} \cos{\frac{2k\pi}{n}} = -\frac{1}{2} \),

where \(n\) is an odd integer, without clearly explaining it. This note explains it.

Without loss of generality, let us use \(n=9\). Then we note that \(\cos{\frac{k\pi}{9}}\), where \(k = 0,2,3,...17\) are the real part of \(\omega^{18}\) the \(18^{th}\) roots of unity, as \(e^{\frac{k\pi}{9}} = \cos{\frac{k\pi}{9}} + i \sin{\frac{k\pi}{9}} \).

Therefore, they can be represented by a Argand diagram (left). It is easy to see from the Argand diagram that:

  • \(\displaystyle \sum_ {k=0}^{17} \cos{\frac{k\pi}{9}} = 0 \) and
  • \(\displaystyle \sum_ {k=0}^{17} \sin{\frac{k\pi}{9}} = 0 \)

because for every root \(\omega^i\) there is an opposite \(-\omega^{9+i}\).

The Argand diagram shows the fact that \(\displaystyle \sum_ {k=0}^{n-1} \omega^k = 0 \), which is true for \(\omega^{n} = 1\).

Now, if we remove the even roots \(k = 0, 2, 4, ... 16\). The Argand diagram (right) is left with 9 uniform distributed odd roots. Does the Argand diagram shows that:

\(\displaystyle \sum_{k=1}^{\color{red}{9}} \omega^{2k-1} = 0 \)?

Yes, it does. We can check it out numerically. But can we explain it without using numerical method? Yes, I have a way which does not resort to serious algebra but classical mechanics.

Let the real and imaginary axes be replaced by \(x\)- and \(y\) axes on a horizontal plane respectively. Let the 9 roots be identical thin solid rod of unit length each with mass \(m\) and are connected firmly at the center, like the spokes of a wheel without a rim. Let the wheel be pivoted freely at the center. We can expect it to be balanced and stay horizontal because the rods are symmetrically distributed from the center. This means that the resultant moment of the wheel on any horizontal axis through the center is \(0\). If we take the moments about \(y\)-axis, then the resultant moment:

\(\displaystyle \sum_{k=1}^{9} \frac{1}{2}\cos{\frac{(2k-1)\pi}{9}}mg = 0 \),

where \(g\) is acceleration due to gravity and the factor \(\frac{1}{2}\) is because center of mass is at the middle of the rod.

This implies that :

  • \(\displaystyle \sum_{k=1}^{9} \cos{\frac{(2k-1)\pi}{9}} = 0 \) and similarly,
  • \(\displaystyle \sum_{k=1}^{9} \sin{\frac{(2k-1)\pi}{9}} = 0 \) and therefore,
  • \(\displaystyle \sum_{k=1}^{9} \omega^{2k-1} = 0 \).

We can conclude that if the complex numbers represented by an Argand diagram is symmetrical from the center the sum of the complex number is \(0\).

We know that \(\cos{\frac{k\pi}{9}} = \cos{\frac{-k\pi}{9}} = \cos{\frac{(18-k)\pi}{9}}\), therefore,

\(\begin{equation} \begin{split} \sum_{k=1}^{9} \cos{\frac{(2k-1)\pi}{9}} & = 0 \\ 2 \sum_{k=1}^{4} \cos{\frac{(2k-1)\pi}{9}} - \cos{\pi }& = 0 \\ 2 \sum_{k=1}^{4} \cos{\frac{(2k-1)\pi}{9}} - 1 & = 0 \\ \Rightarrow \sum_{k=1}^{4} \cos{\frac{(2k-1)\pi}{9}} & = \frac{1}{2} \end{split} \end{equation} \)

Therefore, showing \(\displaystyle \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} \cos{\frac{(2k-1)\pi}{n}} = \frac{1}{2} \), when \(n\) is an odd integer. Similar explanation can be used for \(\displaystyle \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} \cos{\frac{2k\pi}{n}} = -\frac{1}{2} \).

Note by Chew-Seong Cheong
1 year, 6 months ago

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(√2+i√2/3)^(1+i),what is the value of it ? Sumit Ghosh · 1 year, 5 months ago

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@Sumit Ghosh \(\begin{equation} \begin{split} \left( \sqrt{2} + i\frac{\sqrt{2}}{3} \right)^{1+i} & = \left(\frac{\sqrt{20}}{3} \right)^{1+i} \left( \frac{3}{\sqrt{10}} + i\frac{1}{\sqrt{10}} \right)^{1+i} = \left(\frac{\sqrt{20}}{3} \right) \left( e^{i\ln{\frac{\sqrt{20}}{3}}} \right) \left( e^{i \tan^{-1}\frac{1}{3}} \right)^{1+i} \\ & = \frac {\sqrt{20} e^{i\ln{\frac{\sqrt{20}}{3}}} e^{\tan^{-1}\frac{1}{3}(-1+i)}} {3} = \frac {\sqrt{20} e^{i\ln{\frac{\sqrt{20}}{3}}} e^{i\tan^{-1}\frac{1}{3}}} {3e^{\tan^{-1}\frac{1}{3}}} \\ & = \frac {\sqrt{20}e^{i\ln{\frac{\sqrt{20}}{3}}}} {3e^{\tan^{-1}\frac{1}{3}}} \left( \frac{3}{\sqrt{10}} + i\frac{1}{\sqrt{10}} \right) \\ & = \frac {\sqrt{2}\left(\cos{\left(\ln{\frac{\sqrt{20}}{3}}\right)} +i \sin{\left(\ln{\frac{\sqrt{20}}{3}}\right)}\right)(3+i)} {3e^{\tan^{-1}\frac{1}{3}}} \\ & = \boxed{0.811674521 + i0.713337315} \end{split} \end{equation} \) Chew-Seong Cheong · 1 year, 5 months ago

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amazing.great piece of information! :)) Satyabrata Dash · 7 months, 3 weeks ago

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