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$z=\large\frac{1}{1-\cos \theta+i\sin \theta}$.Re(z) is equal to

$\begin{array}{1 1}(A)\large\frac{1}{1-\cos \theta} \\ (B) 1- \cos \theta \\ (C) \large\frac{1}{2} \\(D) 2 \end{array}$

1 Answer

Let $z=x+iy$
$x+iy=\large\frac{1}{1-\cos \theta+i\sin \theta}$
$\Rightarrow \large\frac{(1-\cos \theta)-i\sin \theta}{(1-\cos\theta)^2+\sin^2\theta}$
$\Rightarrow \large\frac{(1-\cos \theta)-i\sin \theta}{2-2\cos\theta}$
$\Rightarrow x=\large\frac{(1-\cos \theta)}{2(1-\cos \theta)}$
$\Rightarrow x=\large\frac{1}{2}$
Hence (C) is the correct answer.
answered Apr 9, 2014 by sreemathi.v
edited Apr 9, 2014 by sreemathi.v
 

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