Invoking Complex Analysis!

Algebra Level 4

If x r = cos ( π 5 r ) + i sin ( π 5 r ) x_r = \cos \left( \dfrac{\pi}{5^r} \right) + i \sin \left( \dfrac{\pi}{5^r} \right) , evaluate the value of:

( r = 1 x r ) + ( r = 1 x r ) + r = 1 x r \large \Re\left(\prod_{r=1}^\infty x_r \right) + \Im\left(\prod_{r=1}^\infty x_r \right) + \left|\prod_{r=1}^\infty x_r \right|

where ( z ) \Re(z) , ( z ) \Im(z) and z |z| are the real part, imaginary part and absolute value of complex number z z .


The answer is 2.414213562.

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1 solution

Kishore S. Shenoy
Sep 22, 2015

x r = e i π n r \large x_r = e^{\frac{i\pi}{n^r}}

r = 1 x r = x 1 x 2 x 3 x 4 = e i π n 1 e i π n 2 e i π n 3 = e i π ( 1 n + 1 n 2 + 1 n 3 + ) = e i π 1 n × 1 1 1 n = e i π × 1 n 1 = cos π n 1 + i sin π n 1 n = 5 , r = 1 = cos π 4 + i sin π 4 = 1 2 + i 2 \displaystyle\begin{aligned}\Rightarrow \prod\limits_{r = 1}^\infty x_r &= x_1x_2x_3x_4\cdots\infty\\ &=e^{\dfrac{i\pi}{n^1}}e^{\dfrac{i\pi}{n^2}}e^{\dfrac{i\pi}{n^3}} \cdots \infty\\ &=e^{i\pi\left(\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\cdots\infty\right)}\\ &=e^{i\pi\dfrac{1}{n}\times\dfrac{1}{1-\frac{1}{n}}}\\ &=e^{i\pi\times\dfrac{1}{n-1}}\\ &=\cos\dfrac{\pi}{n-1}+i\sin\dfrac{\pi}{n-1}\\\\ n = 5, \\ \prod \limits_{r=1}^{\infty} &= \cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\\ &=\dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}\end{aligned}

R e ( r = 1 x r ) + I m ( r = 1 x r ) + r = 1 x r = 2 + 1 2.4142 \Large\therefore \boxed{Re\left(\prod\limits_{r=1}^{\infty} x_r\right) + Im\left(\prod\limits_{r=1}^{\infty} x_r\right) + \left|\prod\limits_{r=1}^{\infty} x_r\right| =\huge\sqrt{2}+1}\Large\approx 2.4142

Moderator note:

Simple standard approach once you recognize the Geometric progression in the powers.

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