2015_7 Complex Indeed! -part 2

Geometry Level 4

∑ m = 1 4 n + 1 { ∑ k = 1 m − 1 { sin ⁡ ( 2 k π m ) − i cos ⁡ ( 2 k π m ) } } m \sum _{ m=1 }^{ 4n+1 }{ { \left\{ \sum _{ k=1 }^{ m-1 }{ \left\{ \sin\left( \frac { 2k\pi }{ m } \right) -i\cos\left( \frac { 2k\pi }{ m } \right) \right\} } \right\} }^{ m } }

where i = − 1 i=\sqrt { -1 } .

What can we describe for the expression above?

More question?!

purely real purely imaginary has imaginary part equal to 2 has real part equal to 1

Anandhu Raj by Anandhu Raj , 23, India

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1 solution

Anandhu Raj
Feb 28, 2015

Given ∑ m = 1 4 n + 1 { ∑ k = 1 m − 1 { s i n ( 2 k π m ) − i c o s ( 2 k π m ) } } m \sum _{ m=1 }^{ 4n+1 }{ { \left\{ \sum _{ k=1 }^{ m-1 }{ \left\{ sin\left( \frac { 2k\pi }{ m } \right) -icos\left( \frac { 2k\pi }{ m } \right) \right\} } \right\} }^{ m } }

Evaluating ∑ k = 1 m − 1 { s i n ( 2 k π m ) − i c o s ( 2 k π m ) } \sum _{ k=1 }^{ m-1 }{ \left\{ sin\left( \frac { 2k\pi }{ m } \right) -icos\left( \frac { 2k\pi }{ m } \right) \right\} }

= ∑ k = 1 m − 1 ( − i ) { c o s ( 2 k π m ) + i s i n ( 2 k π m ) } =\sum _{ k=1 }^{ m-1 }{ (-i)\left\{ cos\left( \frac { 2k\pi }{ m } \right) +isin\left( \frac { 2k\pi }{ m } \right) \right\} }

= ( − i ) ∑ k = 1 m − 1 e 2 π k m =(-i)\sum _{ k=1 }^{ m-1 }{ { e }^{ \frac { 2\pi k }{ m } } }

= ( − i ) ∑ k = 1 m − 1 α k , w h e r e α = e 2 π m =(-i)\sum _{ k=1 }^{ m-1 }{ { \alpha }^{ k } } \quad \quad \quad \quad \quad ,where\quad \alpha ={ e }^{ \frac { 2\pi }{ m } }

= ( − i ) α ( α m − 1 − 1 ) ( α − 1 ) =(-i)\frac { \alpha ({ \alpha }^{ m-1 }-1) }{ (\alpha -1) }

⇒ ∑ k = 1 m − 1 { s i n ( 2 k π m ) − i c o s ( 2 k π m ) } = i \Rightarrow \sum _{ k=1 }^{ m-1 }{ \left\{ sin\left( \frac { 2k\pi }{ m } \right) -icos\left( \frac { 2k\pi }{ m } \right) \right\} } =i

Now ∑ m = 1 4 n + 1 { ∑ k = 1 m − 1 { s i n ( 2 k π m ) − i c o s ( 2 k π m ) } } m = ∑ m = 1 4 n + 1 i m \sum _{ m=1 }^{ 4n+1 }{ { \left\{ \sum _{ k=1 }^{ m-1 }{ \left\{ sin\left( \frac { 2k\pi }{ m } \right) -icos\left( \frac { 2k\pi }{ m } \right) \right\} } \right\} }^{ m } } =\sum _{ m=1 }^{ 4n+1 }{ { i }^{ m } }

⇒ ∑ m = 1 4 n + 1 { ∑ k = 1 m − 1 { s i n ( 2 k π m ) − i c o s ( 2 k π m ) } } m = i = p u r e l y i m a g i n a r y \Rightarrow \sum _{ m=1 }^{ 4n+1 }{ { \left\{ \sum _{ k=1 }^{ m-1 }{ \left\{ sin\left( \frac { 2k\pi }{ m } \right) -icos\left( \frac { 2k\pi }{ m } \right) \right\} } \right\} }^{ m } } =i=\boxed { purely\quad imaginary }

4 Helpful 1 Interesting 0 Brilliant 0 Confused

How is ∑ m = 1 4 n + 1 i m \sum _{ m=1 }^{ 4n+1 }{ { i }^{ m } } = i

Parth Deshpande - 5 years, 6 months ago

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∑ m = 1 4 n + 1 i m = ∑ m = 1 4 n i m + i \sum _{ m=1 }^{ 4n+1 }{ { i }^{ m } } =\sum _{ m=1 }^{ 4n }{ { i }^{ m } } +i

and ∑ m = 1 4 n i m \sum _{ m=1 }^{ 4n }{ { i }^{ m } } is always equal to zero. You can check it yourselves!

Anandhu Raj - 5 years, 6 months ago

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