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Question

# If $z=\frac{1}{1-\mathrm{cos\theta }-i\mathrm{sin\theta }}$, then Re (z) = (a) 0 (b) $\frac{1}{2}$ (c) $\mathrm{cot}\frac{\mathrm{\theta }}{2}$ (d) $\frac{1}{2}\mathrm{cot}\frac{\mathrm{\theta }}{2}$

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Solution

## (b) $\frac{1}{2}$ $z=\frac{1}{1-\mathrm{cos}\theta -i\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}z=\frac{1}{1-\mathrm{cos}\theta -i\mathrm{sin}\theta }×\frac{1-\mathrm{cos}\theta +i\mathrm{sin}\theta }{1-\mathrm{cos}\theta +i\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}⇒z=\frac{1-\mathrm{cos}\theta +i\mathrm{sin}\theta }{{\left(1-\mathrm{cos}\theta \right)}^{2}-{\left(i\mathrm{sin}\theta \right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒z=\frac{1-\mathrm{cos}\theta +i\mathrm{sin}\theta }{1+{\mathrm{cos}}^{2}\theta -2\mathrm{cos}\theta +{\mathrm{sin}}^{2}\theta }\phantom{\rule{0ex}{0ex}}⇒z=\frac{1-\mathrm{cos}\theta +i\mathrm{sin}\theta }{1+1-2\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}⇒z=\frac{1-\mathrm{cos}\theta +i\mathrm{sin}\theta }{2\left(1-\mathrm{cos}\theta \right)}\phantom{\rule{0ex}{0ex}}⇒\mathrm{R}\mathrm{e}\left(z\right)=\frac{\left(1-\mathrm{cos}\theta \right)}{2\left(1-\mathrm{cos}\theta \right)}=\frac{1}{2}$

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