Complex or simple?
I f z = e i θ t h e n f i n d ∑ n = 1 1 5 I m ( z 2 n − 1 ) a t θ = 3 0 ∘ The question is not original.
The answer is 2.
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1 solution
Generalising
sin α + sin ( α + β ) + s i n ( α + 2 β ) + … sin ( α + ( n − 1 ) β ) = sin ( 2 β ) sin ( 2 n β ) ( sin ( α + ( n − 1 ) 2 β )
z = e i θ ⟹ z 2 n − 1 = e i ∗ ( 2 n − 1 ) ∗ θ t h e r e f o r e , I m ( z 2 n − 1 ) = sin ( 2 n − 1 ) ∗ θ ⟹ ∑ n = 1 1 5 I m ( z 2 n − 1 ) = sin θ + sin 3 θ + sin 5 θ + . . . . + sin 2 9 θ = S . T h e r e f o r e , S ∗ sin θ = sin 2 θ + sin 3 θ ∗ sin θ + sin 5 θ ∗ sin θ + . . . . . . sin 2 9 θ ∗ sin θ ⟹ S ∗ 2 sin θ = ( 1 − cos 2 θ ) + ( cos 2 θ − cos 4 θ + . . . . . ( cos 2 8 θ − cos 3 0 θ ) ⟹ S ∗ 2 sin θ = 1 − cos 3 0 θ . T h e r e f o r e , S = 2 sin θ 1 − cos 3 0 θ ⟹ S = 2 sin 3 0 ∘ 1 − cos 9 0 0 ∘ = 2