AP EAMCET · Maths · Complex Number
Let \(z=\cos \theta+i \sin \theta\). Then, the value of \(\sum_{m=1}^{15} I_m\left(z^{2 m-1}\right)\) at \(\theta=2^{\circ}\) is
- A \(\frac{1}{\sin 2^{\circ}}\)
- B \(\frac{1}{3 \sin 2^{\circ}}\)
- C \(\frac{1}{2 \sin 2^{\circ}}\)
- D \(\frac{1}{4 \sin 2^{\circ}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{4 \sin 2^{\circ}}\)
Step-by-step Solution
Detailed explanation
\(z=\cos \theta+i \sin \theta=e^{i \theta}\) [in Euler's form] Using Demoivre theorem \[ \begin{aligned} z^2 & =\cos 2 \theta+i \sin 2 \theta \\ z^3 & =\cos 3 \theta+i \sin 3 \theta \end{aligned} \] Now, according to the question…
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