Argand Diagram and k0n1cos(2k+12n+1π)=12\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:

  • k=1n2cos(2k1)πn=12\displaystyle \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} \cos{\frac{(2k-1)\pi}{n}} = \frac{1}{2} and
  • k=1n2cos2kπn=12\displaystyle \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} \cos{\frac{2k\pi}{n}} = -\frac{1}{2} ,

where nn is an odd integer, without clearly explaining it. This note explains it.

Without loss of generality, let us use n=9n=9. Then we note that coskπ9\cos{\frac{k\pi}{9}}, where k=0,2,3,...17k = 0,2,3,...17 are the real part of ω18\omega^{18} the 18th18^{th} roots of unity, as ekπ9=coskπ9+isinkπ9e^{\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:

  • k=017coskπ9=0\displaystyle \sum_ {k=0}^{17} \cos{\frac{k\pi}{9}} = 0 and
  • k=017sinkπ9=0\displaystyle \sum_ {k=0}^{17} \sin{\frac{k\pi}{9}} = 0

because for every root ωi\omega^i there is an opposite ω9+i-\omega^{9+i}.

The Argand diagram shows the fact that k=0n1ωk=0\displaystyle \sum_ {k=0}^{n-1} \omega^k = 0 , which is true for ωn=1\omega^{n} = 1.

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

k=19ω2k1=0\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 xx- and yy axes on a horizontal plane respectively. Let the 9 roots be identical thin solid rod of unit length each with mass mm 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 00. If we take the moments about yy-axis, then the resultant moment:

k=1912cos(2k1)π9mg=0\displaystyle \sum_{k=1}^{9} \frac{1}{2}\cos{\frac{(2k-1)\pi}{9}}mg = 0 ,

where gg is acceleration due to gravity and the factor 12\frac{1}{2} is because center of mass is at the middle of the rod.

This implies that :

  • k=19cos(2k1)π9=0\displaystyle \sum_{k=1}^{9} \cos{\frac{(2k-1)\pi}{9}} = 0 and similarly,
  • k=19sin(2k1)π9=0\displaystyle \sum_{k=1}^{9} \sin{\frac{(2k-1)\pi}{9}} = 0 and therefore,
  • k=19ω2k1=0\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 00.

We know that coskπ9=coskπ9=cos(18k)π9\cos{\frac{k\pi}{9}} = \cos{\frac{-k\pi}{9}} = \cos{\frac{(18-k)\pi}{9}}, therefore,

k=19cos(2k1)π9=02k=14cos(2k1)π9cosπ=02k=14cos(2k1)π91=0k=14cos(2k1)π9=12 \begin{aligned} \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{aligned}

Therefore, showing k=1n2cos(2k1)πn=12\displaystyle \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} \cos{\frac{(2k-1)\pi}{n}} = \frac{1}{2} , when nn is an odd integer. Similar explanation can be used for k=1n2cos2kπn=12\displaystyle \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} \cos{\frac{2k\pi}{n}} = -\frac{1}{2} .

Note by Chew-Seong Cheong
3 years, 11 months ago

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(√2+i√2/3)^(1+i),what is the value of it ?

Sumit Ghosh - 3 years, 10 months ago

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(2+i23)1+i=(203)1+i(310+i110)1+i=(203)(eiln203)(eitan113)1+i=20eiln203etan113(1+i)3=20eiln203eitan1133etan113=20eiln2033etan113(310+i110)=2(cos(ln203)+isin(ln203))(3+i)3etan113=0.811674521+i0.713337315 \begin{aligned} \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{aligned}

Chew-Seong Cheong - 3 years, 10 months ago

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amazing.great piece of information! :))

Satyabrata Dash - 3 years ago

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