Question

The real part of $\frac{1}{1+Cos\theta +iSin\theta }$ is

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (B)

Let $z=\frac{1}{1+\cos \theta +i\sin \theta }$
$=\frac{1}{2{\cos }^2\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}$
$=\frac{1}{2\cos \frac{\theta }{2}\left(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}\right)}$
$=\frac{1}{2\cos \frac{\theta }{2}}{\left(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}\right)}^{-1}$
$=\frac{\cos \left(-\frac{\theta }{2}\right)+i\sin \left(-\frac{\theta }{2}\right)}{2\cos \frac{\theta }{2}}=\frac{\cos \frac{\theta }{2}-i\sin \frac{\theta }{2}}{2\cos \frac{\theta }{2}}$
[using De-Moivre's Theorem]
$=\frac{1}{2}\left[1-i\tan \frac{\theta }{2}\right]$
Real of $z=\frac{1}{2}$ real part of $\left[1-i\tan \frac{\theta }{2}\right]$
$=\frac{1}{2}(1)=\frac{1}{2}$