Let z = cos + i sin . Then the value of m = 1 15 Im ( z 2 m - 1 ) at = 2 is [2009]

Question

Let $z = \cos θ + i \sin θ$ . Then the value of $\sum_{m = 1}^{15} Im (z^{2 m - 1})$ at $θ = 2 °$ is [2009]

Smart Achievers · Accepted Answer

(4)

$z = \cos θ + i \sin θ$

$\Rightarrow z^{2 m - 1} = {(\cos θ + i \sin θ)}^{2 m - 1} = \cos (2 m - 1) θ + i \sin (2 m - 1) θ$

$[By De Moivre's theorem: {(\cos θ + i \sin θ)}^{n} = \cos n θ + i \sin n θ]$

$∴$ $Im (z^{2 m - 1}) = \sin (2 m - 1) θ$

$∴$ $\sum_{m = 1}^{15} Im (z^{2 m - 1}) = \sum_{m = 1}^{15} \sin (2 m - 1) θ$

$= \sin θ + \sin 3 θ + \sin 5 θ + \dots (upto 15 terms)$

$= \frac{\sin [15 (\frac{2 θ}{2})] \cdot \sin [θ + 14 θ]}{\sin θ}$

$[\begin{matrix} ∵ \sin α + \sin (α + β) + \sin (α + 2 β) + \dots (n terms) \\ = \frac{\sin (\frac{n β}{2}) \cdot \sin (α + \frac{(n - 1) β}{2})}{\sin (\frac{β}{2})} \end{matrix}]$

$= \frac{\sin 15 θ \cdot \sin 15 θ}{\sin θ} = \frac{\sin 30 ° \cdot \sin 30 °}{\sin 2 °} = \frac{1}{4 \sin 2 °}$

Solution

Q.
Let $z = \cos θ + i \sin θ$ . Then the value of $\sum_{m = 1}^{15} Im (z^{2 m - 1})$ at $θ = 2 °$ is [2009]

1 $\frac{1}{\sin 2 °}$

2 $\frac{1}{3 \sin 2 °}$

3 $\frac{1}{2 \sin 2 °}$

4 $\frac{1}{4 \sin 2 °}$

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Solution

Q. Let z=cosθ+isinθ. Then the value of ∑m=115Im(z2m-1) at θ=2° is [2009]

1 1sin2°

2 13sin2°

3 12sin2°