If z = cos 6° + i sin 6° , then ₙ=1}²0} {Im}(z²n-1}) =

Correct Answer is : 34sin⁡6∘{3}{4 6^{}}4sin6∘3

Correct Answer is : 34sin⁡6∘{3}{4 6^{}}4sin6∘3

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AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Section: Mathematics

Question : 10 of 160

Marks: +1, -0

\sum\limits_{n=1}^{20} \operatorname{Im}(z^{2n-1})

Solution:

\sum\limits_{n=1}^{20} z^{2n-1}=z+z^{3}+z^{5}+\dots+z^{39}

\text{Now}, z=\cos 6^{\circ}+i \sin 6^{\circ}=e^{i 6^{\circ}}

Substitute the value in the series,

\sum\limits_{n=1}^{20} z^{2n-1} \; = e^{i 6^{\circ}}+e^{i 18^{\circ}}+e^{i 30^{\circ}}+\dots+e^{i(39 \times 6^{\circ})}

\; = e^{i 6^{\circ}}(1+e^{i 12}+e^{i 24}+\dots+e^{i 228})

\; = e^{i 6^{\circ}} \left[ \frac{e^{i 240}-1}{e^{i 12}-1} \right]

Simplify the above equation,

\sum\limits_{n=1}^{20} \operatorname{Im}(z^{2n-1}) \; = e^{i 6^{\circ}} \left[ \frac{e^{i 240}-1}{e^{i 12}-1} \right]

\; = \frac{e^{i 6^{\circ}} \left[ \frac{-1}{2} - \frac{\sqrt{3}}{2} i - 1 \right]}{e^{i 6^{\circ}} (2 i \sin 6^{\circ})}

\; = \frac{ \left[ -\frac{3}{2} - \frac{\sqrt{3}}{2} i \right] }{ 2 i \sin 6^{\circ} }

\; = -\frac{\sqrt{3}}{4 \sin 6^{\circ}} + \frac{\sqrt{3}}{4 \sin 6^{\circ}} i

\text{Thus}, \sum\limits_{n=1}^{20} \operatorname{Im}(z^{2n-1}) \; = \frac{3}{4 \sin 6^{\circ}}

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