Question

Represent $\frac{1}{1-\cos \theta +2i\sin \theta }$ in the form $A+iB$

Answer 1

Rationalizing the denominator of $\frac{1}{1-\cos \theta +2i\sin \theta }$.

$\frac{1}{1-\cos \theta +2i\sin \theta }\times \frac{1-\cos \theta -2i\sin \theta }{1-\cos \theta -2i\sin \theta }$

$=\frac{\left(1-\cos \theta \right)-2i\sin \theta }{{\left(1-\cos \theta \right)}^2-{\left(2i\sin \theta \right)}^2}$

$=\frac{\left(1-\cos \theta \right)-2i\sin \theta }{{\left(1-\cos \theta \right)}^2+4{\sin }^2\theta }$

$=\frac{\left(1-\cos \theta \right)-2i\sin \theta }{1+{\cos }^2\theta -2\cos \theta +4\left(1-{\cos }^2\theta \right)}$

$=\frac{\left(1-\cos \theta \right)-2i\sin \theta }{1+{\cos }^2\theta -2\cos \theta +4-4{\cos }^2\theta }$

$=\frac{\left(1-\cos \theta \right)-2i\sin \theta }{-3{\cos }^2\theta -2\cos \theta +5}$

$=\frac{\left(1-\cos \theta \right)-2i\sin \theta }{-3{\cos }^2\theta -5\cos \theta +3\cos \theta +5}$

$=\frac{\left(1-\cos \theta \right)-2i\sin \theta }{-\cos \theta \left(3\cos \theta +5\right)+1\left(3\cos \theta +5\right)}$

$=\frac{\left(1-\cos \theta \right)-2i\sin \theta }{\left(3\cos \theta +5\right)\left(1-\cos \theta \right)}$

$=\frac{1-\cos \theta }{\left(3\cos \theta +5\right)\left(1-\cos \theta \right)}-\frac{2i\sin \theta }{\left(3\cos \theta +5\right)\left(1-\cos \theta \right)}$

$=\frac{1}{3\cos \theta +5}+i\frac{2\sin \theta }{\left(3\cos \theta +5\right)\left(\cos \theta -1\right)}$