Demoive formula

Algebra Level pending

( 2 + i 5 2 − i 5 ) 10 + ( 2 − i 5 2 + i 5 ) 10 \left(\frac{2+i\sqrt{5}}{2-i\sqrt{5}}\right)^{10}+ \left(\frac{2-i\sqrt{5}}{2+i\sqrt{5}}\right)^{10}

Find the modulus of the complex number above.

2 cos ⁡ ( 10 cos ⁡ − 1 2 3 ) 2\cos \left(10\cos^{-1} \frac{2}{3}\right) 2 cos ⁡ ( 20 cos ⁡ − 1 2 3 ) 2\cos \left(20\cos^{-1} \frac{2}{3}\right) 2 sin ⁡ ( 10 cos ⁡ − 1 2 3 ) 2\sin \left(10\cos^{-1} \frac{2}{3}\right) 2 sin ⁡ ( 20 cos ⁡ − 1 2 3 ) 2\sin \left(20\cos^{-1} \frac{2}{3}\right)

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Chew-Seong Cheong
May 13, 2020

z = ( 2 + 5 i 2 − 5 i ) 10 + ( 2 − 5 i 2 + 5 i ) 10 = ( ( 2 + 5 i ) 2 9 ) 10 + ( ( 2 − 5 i ) 2 9 ) 10 = ( − 1 + 4 5 i 9 ) 10 + ( − 1 − 4 5 i 9 ) 10 = e 10 cos ⁡ − 1 ( − 1 9 ) i + e − 10 cos ⁡ − 1 ( − 1 9 ) i By Euler’s formula: e θ i = cos ⁡ θ + i sin ⁡ θ = 2 cos ⁡ ( 10 cos ⁡ − 1 ( − 1 9 ) ) Since 2 cos ⁡ − 1 2 3 = cos ⁡ − 1 ( − 1 9 ) = 2 cos ⁡ ( 20 cos ⁡ − 1 2 3 ) \begin{aligned} z & = \left(\frac {2 + \sqrt 5 i}{2 - \sqrt 5 i} \right)^{10} + \left(\frac {2 - \sqrt 5 i}{2 + \sqrt 5 i} \right)^{10} \\ & = \left(\frac {(2 + \sqrt 5 i)^2}9 \right)^{10} + \left(\frac {(2 - \sqrt 5 i)^2}9 \right)^{10} \\ & = \left(\frac {-1 + 4\sqrt 5 i}9 \right)^{10} + \left(\frac {-1-4\sqrt5 i}9 \right)^{10} \\ & = \large e^{10\cos^{-1}\left(-\frac 19\right)i} + e^{-10 \cos^{-1}\left(-\frac 19\right)i} & \small \blue{\text{By Euler's formula: }e^{\theta i} = \cos \theta + i \sin \theta} \\ & = 2 \cos \left(10\cos^{-1}\left(-\frac 19\right)\right) & \small \blue{\text{Since }2 \cos^{-1} \frac 23 = \cos^{-1} \left(- \frac 19 \right)} \\ & = 2 \cos \left(20\cos^{-1}\frac 23 \right) \end{aligned}

Therefore, ∣ z ∣ = 2 cos ⁡ ( 20 cos ⁡ − 1 2 3 ) |z| = \boxed{2 \cos \left(20\cos^{-1}\frac 23 \right)} .

Reference: Euler's formula

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The given expression is

e i ( 20 tan ⁡ − 1 5 2 ) + e − i ( 20 tan ⁡ − 1 5 2 ) e^{i(20\tan^{-1} \frac{\sqrt 5}{2})}+e^{-i(20\tan^{-1} \frac{\sqrt 5}{2})}

= 2 cos ⁡ ( 20 tan ⁡ − 1 5 2 ) = 2 cos ⁡ ( 20 cos ⁡ − 1 2 3 ) =2\cos (20\tan^{-1} \frac{\sqrt 5}{2})=\boxed {2\cos \left (20\cos^{-1} \frac{2}{3}\right )}

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change t a n − 1 ( 5 2 ) tan^{-1}\big(\frac{\sqrt{5}}{2} \big) into c o s − 1 2 3 cos^{-1} \frac{2}{3}

A Former Brilliant Member - 1 year, 1 month ago

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Yeah. Edited.

A Former Brilliant Member - 1 year, 1 month ago

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