Question

The expression of the complex number $\frac{1}{1+\cos \theta -i\sin \theta }$ in the form $a+ib$ is

Answer 1

Rationalize the given complex number.

A complex number $\frac{1}{1+\cos \theta -i\sin \theta }$ is given.

Multiply and divide the given complex number with the conjugate of the denominator.

$$\begin{gathered} \frac{1}{1+\cos \theta -i\sin \theta }=\frac{1}{1+\cos \theta -i\sin \theta }·\frac{1+\cos \theta +i\sin \theta }{1+\cos \theta +i\sin \theta } \\ \Rightarrow \frac{1}{1+\cos \theta -i\sin \theta }=\frac{1+\cos \theta +i\sin \theta }{{\left(1+\cos \theta \right)}^2-{\left(i\sin \theta \right)}^2} \\ \Rightarrow \frac{1}{1+\cos \theta -i\sin \theta }=\frac{1+\cos \theta +i\sin \theta }{1+{\cos }^2\theta +2\cos \theta +{\sin }^2\theta } \\ \Rightarrow \frac{1}{1+\cos \theta -i\sin \theta }=\frac{1+\cos \theta +i\sin \theta }{1+1+2\cos \theta } \\ \Rightarrow \frac{1}{1+\cos \theta -i\sin \theta }=\frac{1+\cos \theta +i\sin \theta }{2+2\cos \theta } \\ \Rightarrow \frac{1}{1+\cos \theta -i\sin \theta }=\frac{1+\cos \theta +i\sin \theta }{2\left(1+\cos \theta \right)} \\ \Rightarrow \frac{1}{1+\cos \theta -i\sin \theta }=\frac{1+\cos \theta }{2\left(1+\cos \theta \right)}+\frac{i\sin \theta }{2\left(1+\cos \theta \right)} \\ \Rightarrow \frac{1}{1+\cos \theta -i\sin \theta }=\frac{1}{2}+\frac{2·i·\cos \left({\displaystyle \frac{\theta }{2}}\right)·\sin \left({\displaystyle \frac{\theta }{2}}\right)}{2\left(2{\cos }^2\left(\frac{\theta }{2}\right)\right)} \\ \Rightarrow \frac{1}{1+\cos \theta -i\sin \theta }=\frac{1}{2}+\frac{i·\sin \left({\displaystyle \frac{\theta }{2}}\right)}{2\cos \left(\frac{\theta }{2}\right)} \\ \Rightarrow \frac{1}{1+\cos \theta -i\sin \theta }=\frac{1}{2}+i·\tan \left(\frac{\theta }{2}\right) \end{gathered}$${Since, $1+\cos \theta =2{\cos }^2\left(\frac{\theta }{2}\right)$}

Therefore, the given complex number $\frac{1}{1+\cos \theta -i\sin \theta }$ in the form $a+ib$ is $\frac{1}{2}+i·\tan \left(\frac{\theta }{2}\right)$.