Question

If $Z=\cos \theta +i\sin \theta$ find the complex representation of $\frac{Z}{1-2Z}$.

Answer 1

Solution:-

$Z=\cos \theta +i\sin \theta$

$\therefore \frac{Z}{1-2Z}=\frac{\cos \theta +i\sin \theta }{1-2(\cos \theta +i\sin \theta )}$

$\Rightarrow =\frac{\cos \theta +i\sin \theta }{(1-2\cos \theta )-2i\sin \theta }$

Multiplying and dividing by the conjugate of denominator, we have

$\frac{Z}{1-2Z}=\frac{\cos \theta +i\sin \theta }{(1-2\cos \theta )-2i\sin \theta }\times \frac{(1-2\cos \theta )+2i\sin \theta }{(1-2\cos \theta )+2i\sin \theta }$

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

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

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

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

$\Rightarrow =\frac{(\cos \theta -2)+i\sin \theta }{5-4\cos \theta }$

Hence, the complex representation of $\frac{Z}{1-2Z}$ will be $\frac{(\cos \theta -2)+i\sin \theta }{5-4\cos \theta }$.