Let z-cos-i sin- The value of -m-115Imz2 m-1 at -2-

Z=cosθ +i sinθ =e^iθ ∴∑ ^15_m=1 Im(z^2m 1)=∑ ^15_m=1 Im(e^iθ)^2m 1 =∑ ^15_m=1 Im e^i(2m 1)θ =sin θ +sin 3θ +sin 5θ +...+sin 29θ =sin (θ `+ 29θ/2 )sin (15× 2θ ...

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Byju's Answer

Standard XII

Mathematics

Euler's Representation

Let z=cosθ+i ...

Question

Let $z = c o s θ + i s i n θ$ . The value of $\sum_{m = 1}^{15} I m (z^{2 m - 1})$ at $θ = 2^{\circ}$

\frac{1}{s i n 2^{\circ}}

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\frac{1}{3 s i n 2^{\circ}}

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\frac{1}{2 s i n 2^{\circ}}

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\frac{1}{4 s i n 2^{\circ}}

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Solution

The correct option is D $\frac{1}{4 s i n 2^{\circ}}$
$z = c o s θ + i s i n θ = e^{i θ} ∴ \sum_{m = 1}^{15} I m (z^{2 m - 1}) = \sum_{m = 1}^{15} I m (e^{i θ})^{2 m - 1} = \sum_{m = 1}^{15} I m e^{i (2 m - 1) θ} = s i n θ + s i n 3 θ + s i n 5 θ + . . . + s i n 29 θ = \frac{s i n (\frac{θ ‘ + 29 θ}{2}) s i n (\frac{15 \times 2 θ}{2})}{s i n (\frac{2 θ}{2})} = \frac{s i n 15 θ . s i n 15 θ}{s i n θ} = \frac{1}{4 s i n 2^{\circ}}$

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Let z=cosθ+i sinθ. The value of ∑15m=1Im(z2m−1) at θ=2∘

The correct option is D 14sin2∘z=cosθ+i sinθ=eiθ∴∑15m=1Im(z2m−1)=∑15m=1Im(eiθ)2m−1=∑15m=1Im ei(2m−1)θ=sin θ+sin 3θ+sin 5θ+...+sin 29θ=sin(θ‘+ 29θ2)sin(15×2θ2)sin(2θ2)=sin 15θ.sin 15θsin θ=14 sin2∘

Let $z = c o s θ + i s i n θ$ . The value of $\sum_{m = 1}^{15} I m (z^{2 m - 1})$ at $θ = 2^{\circ}$