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Algebra Level 2

e i θ e 2 i θ e 3 i θ . . . e n i θ = 1 \large e ^{ i\theta }e^{ 2i\theta} e^{ 3i\theta }...\large e^{ ni\theta } = 1

Find the value of θ \theta satisfying the above equation for m ∈ N m\in\mathbb{N} .

2 m π n 2 \frac { 2m\pi }{ { n }^{ 2 } } 2 n π m ( m + 1 ) \frac { 2n\pi }{ m(m+1) } m π n 2 \frac { m\pi }{ { n }^{ 2 } } m π n ( n + 1 ) \frac { m\pi }{ n(n+1) } 2 m π n ( n + 1 ) \frac { 2m\pi }{ n(n+1) } 4 m π n ( n + 1 ) \frac { 4m\pi }{ n(n+1) } n π m 2 \frac { n\pi }{ { m }^{ 2 } } 4 m π n 2 \frac { 4m\pi }{ { n }^{ 2 } }

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Saurav Pal by Saurav Pal , 26, India

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2 solutions

1 = e 2 m π i 1 = e^{2m\pi i}

L . H . S = e i × θ × ( n ( n + 1 ) 2 ) = 1 = e 2 m π i L.H.S = e^{i\times \theta \times (\frac{n(n+1)}{2})} = 1 = e^{2m\pi i}

θ = 4 m π n ( n + 1 ) \theta = \frac{4m\pi}{n(n+1)}

11 Helpful 0 Interesting 0 Brilliant 0 Confused
Saurav Pal
Feb 20, 2015

e i θ . e 2 i θ ⋯ e n i θ = ( cos ⁡ θ + i sin ⁡ θ ) ( cos ⁡ 2 θ + i sin ⁡ 2 θ ) ⋯ ( cos ⁡ n θ + i sin ⁡ n θ ) { e }^{ i\theta }.{ e }^{ 2i\theta } \cdots { e }^{ ni\theta }= (\cos\theta + i \sin\theta )(\cos2\theta+i \sin2\theta ) \cdots (\cos n\theta + i \sin n\theta )

Using De Moivre's Theorem, the above expression can be written as

( cos ⁡ θ + i sin ⁡ θ ) 1 + 2 + 3 + ⋯ + n = ( cos ⁡ θ + i sin ⁡ θ ) n ( n + 1 ) 2 = cos ⁡ ( n ( n + 1 ) 2 θ ) + i sin ⁡ ( n ( n + 1 ) 2 θ ) { (\cos\theta +i \sin\theta ) }^{ 1+2+3+\cdots+n } \\ = { { (\cos\theta + i \sin\theta ) } }^{ \frac { n(n+1) }{ 2 } }\\ =\cos \left( \frac { n(n+1) }{ 2 } \theta \right) +i \sin \left( \frac { n(n+1) }{ 2 } \theta \right)

which is equal to 1 (as given in the statement of the problem) and further we can write 1 as ( c o s 0 + i sin ⁡ 0 ) (cos 0 +i \sin 0) .

Therefore we have cos ⁡ ( n ( n + 1 ) 2 θ ) + i sin ⁡ ( n ( n + 1 ) 2 θ ) ⇒ c o s ( n ( n + 1 ) 2 θ ) = c o s 0 n ( n + 1 ) 2 θ = 2 m π ± 0. \cos \left( \frac { n(n+1) }{ 2 } \theta \right) +i \sin \left( \frac { n(n+1) }{ 2 } \theta \right) \\ \Rightarrow cos \left( \frac { n(n+1) }{ 2 } \theta \right) = cos 0 \\ \frac { n(n+1) }{ 2 } \theta = 2m\pi \pm 0.

Hence θ = 4 m π n ( n + 1 ) . □ \theta =\frac { 4m\pi }{ n(n+1) } .\square

10 Helpful 0 Interesting 1 Brilliant 0 Confused

The problem statement should not mention "m" at all. The parameter "m" is introduced during the process of solution in order to index the multiple values of theta that will satisfy the equation.

Doug Gwyn - 4 years, 11 months ago

Why can't we do sin x =0 and x=mπ Thus we get ans just half of original

Kumar Krish - 2 years, 5 months ago

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