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$\large{\begin{cases} P = \dfrac{\tan^{-1}(\alpha)}{\alpha} + \dfrac{\tan^{-1}(\beta)}{\beta} + \dfrac{\tan^{-1}(\gamma)}{\gamma} \\ \text{ . } \\ \text{ . } \\ Q = \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{6n+1} \end{cases} }$

If $\alpha, \ \beta, \ \gamma $ are the three cube roots of unity, submit the value of $\dfrac{P}{Q}$ as your answer.

This was the question I came across today.I know most of the properties of cube roots of unity.But how to handle the arctangents of complex numbers?I never saw such an expression.Any suggestions?

My knowledge of maths is upto high school.So please give your answer accordingly.Thanks.

  • All your questions seem to be got from brilliant.org, the purpose of question is to answer them without "cheating" – Oussama Boussif Sep 11 '15 at 08:12
  • @OussamaBoussif I have no interest in answering the questions on brilliant.org...I simply use it for self practise.You might even see my profile there...i hardly answer any questions there to increase my reputation...that can't be called "cheating" because i dont answer questions there. –  Sep 11 '15 at 08:15
  • That was just on observation I made. But in future questions at least show what you tried, I am sure that you'll be able to solve this one. – Oussama Boussif Sep 11 '15 at 08:20
  • I can calculate Q @OussamaBoussif but not sure about P.....because that involves arctan(w) and arctanc(w^2)....I dont know how to handle that... –  Sep 11 '15 at 08:22
  • You don't need to know anything about inverse trig functions of complex numbers. You do need to know about the Taylor expansion of the arctan, and once you write that down, the answer should be clear. – Ron Gordon Sep 11 '15 at 08:23
  • but is taylor series valid for complex numbers @RonGordon ? I thought of that once.. –  Sep 11 '15 at 08:25
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    They are valid for all numbers of magnitude less than or equal to one. Find a proof of that if you need, but that is the case. – Ron Gordon Sep 11 '15 at 08:28
  • @RonGordon thanks for that...got the proof here http://math.stackexchange.com/questions/29649/why-is-arctanx-x-x3-3x5-5-x7-7-dots –  Sep 11 '15 at 08:32
  • @OussamaBoussif how did you get that formula :O!!!Please let me know!Taylor series?Or some other method? –  Sep 11 '15 at 08:34
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    Just use the fact that $\arctan{x}=\int \frac{1}{1+x^2} dx$ and that $\frac{1}{1+x^2}=\frac{1}{2}(\frac{1}{1-ix}+\frac{1}{1+ix})$ – Oussama Boussif Sep 11 '15 at 08:36
  • @OussamaBoussif nice one...:)...good to learn something new!thanks! –  Sep 11 '15 at 08:38
  • You're welcome. But do you know how to handle the logarithm of complex numbers? – Oussama Boussif Sep 11 '15 at 08:39
  • @OussamaBoussif like $ln z=ln|z|+i arg(z)$ ? –  Sep 11 '15 at 08:42
  • Ah in that case you know it. Now apply the identity and see what you can get from it – Oussama Boussif Sep 11 '15 at 08:47
  • Sorry I did a mistake in that identity here's the correct one: $\arctan{x}=\frac{1}{2i}\ln{\frac{1+ix}{1-ix}}$ – Oussama Boussif Sep 11 '15 at 08:48
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    @OussamaBoussif got the answer using taylor series..its coming 1...now trying your method –  Sep 11 '15 at 08:52

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