Question

If $ab=1+\sqrt{1-b^2}andz=e^{i\theta },then\frac{b}{2a}(1+az)(1+\frac{a}{z})$ is

• A. $1+asin\theta$
• B. $1+bcos\theta$
• C. $bcos\theta$
• D. $asin\theta$

Answer 1

Answer: (B)

$1+bcos\theta$

We have $(ab-1{)}^2=1-b^2$
$\Rightarrow$ $b(1+a^2)=2a$
$\Rightarrow$ $\frac{b}{2a}=\frac{1}{1+a^2}$
Thus, $\frac{b}{2a}(1+az)(1+\frac{a}{z})=\frac{b}{2a}\{1+a^2+a(z+\frac{1}{z})\}$
$=\frac{1}{1+a^2}\{1+a^2+2a\cos \theta \}$
$=1+\frac{2a}{1+a^2}\cos \theta =1+b\cos \theta$