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If Z = 1 1 − C O S θ − I S I N θ Then Re (Z) =

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Question

If \[z = \frac{1}{1 - cos\theta - i sin\theta}\] then Re (z) =

Options

  • 0

  • \[\frac{1}{2}\]

  • \[\cot\frac{\theta}{2}\]

  • \[\frac{1}{2}\cot\frac{\theta}{2}\]

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Solution

\[\frac{1}{2}\]

\[z = \frac{1}{1 - \cos\theta - i\sin\theta}\]

\[z = \frac{1}{1 - \cos\theta - i\sin\theta} \times \frac{1 - \cos\theta + i\sin\theta}{1 - \cos\theta + i\sin\theta}\]

\[ \Rightarrow z=\frac{1 - \cos\theta + i\sin\theta}{\left( 1 - \cos\theta \right)^2 - \left( i\sin\theta \right)^2}\]

\[ \Rightarrow z=\frac{1 - \cos\theta + i\sin\theta}{1 + \cos^2 \theta - 2\cos\theta + \sin^2 \theta}\]

\[ \Rightarrow z= \frac{1 - \cos\theta + i\sin\theta}{1 + 1 - 2\cos\theta}$\]

\[ \Rightarrow z=\frac{1 - \cos\theta + i\sin\theta}{2(1 - \cos\theta)}\]

\[ \Rightarrow \text { Re }(z)=\frac{\left( 1 - \cos\theta \right)}{2\left( 1 - \cos\theta \right)}=\frac{1}{2}\]

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Chapter 13: Complex Numbers - Exercise 13.6 [Page 65]

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R.D. Sharma Mathematics [English] Class 11
Chapter 13 Complex Numbers
Exercise 13.6 | Q 25 | Page 65

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