Question

The modulus and argument of ${\left(\frac{1+\cos \theta +i\sin \theta }{1+\cos \theta -i\sin \theta }\right)}^n$ are represented by the set,

• A. $\{2,n\theta \}$
• B. $\{1,n\theta \}$
• C. $\{1,(n-1)\theta \}$
• D. none of these

Answer 1

The correct option is B $\{1,n\theta \}$

$z={\left(\frac{1+\cos \theta +i\sin \theta }{1+\cos \theta -i\sin \theta }\right)}^n\Rightarrow z={\left(\frac{2{\cos }^2\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}-2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right)}^n\Rightarrow z={\left(\frac{\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-i\sin \frac{\theta }{2}}\right)}^n={(\cos \theta +i\sin \theta )}^n\Rightarrow z=\cos n\theta +i\sin n\theta$
$\therefore \left|z\right|=1andargz=n\theta$
Ans: B