Question

If $\frac{3}{(2+\cos \theta +i\sin \theta )}=a+ib$, then $(a-2{)}^2+b^2$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Step 1. Find the value of $(a-2{)}^2+b^2$:

Given, $\frac{3}{(2+\cos \theta +i\sin \theta )}=a+ib$

Rationalize it,

$\begin{array}{rcl}\frac{3}{2+\cos \theta +i\sin \theta }& =& \frac{3}{2+\cos \theta +i\sin \theta }\times \frac{(2+\cos \theta )-i\sin \theta }{(2+\cos \theta )-i\sin \theta }\end{array}$

$\begin{array}{rcl}& =& \frac{(6+3\cos \theta )-i\left(3\sin \theta \right)}{{(2+\cos \theta )}^2+{\sin }^2\theta }\\ & =& \frac{(6+3\cos \theta )-i(3\sin \theta )}{5+4\cos \theta }\end{array}$

Step 2. Compare it with $a+ib$, we get

$a=\frac{6+3\sin \theta }{5+4\cos \theta }$

$b=\frac{-3\sin \theta }{5+4\cos \theta }$

$\therefore$${(a-2)}^2+b^2={\left(\frac{6+3\sin \theta }{5+4\cos \theta }-2\right)}^2+{\left(\frac{-3\sin \theta }{5+4\cos \theta }\right)}^2$

$=\frac{{\left(6+3\sin \theta -2(5+4\cos \theta \right)}^2}{{\left(5+4\cos \theta \right)}^2}+\frac{{\left(-3\sin \theta \right)}^2}{{\left(5+4\cos \theta \right)}^2}$

$$\begin{gathered} =\frac{{(-4-5\cos \theta )}^2+9{\sin }^2\theta }{{(5+4\cos \theta )}^2} \\ =\frac{{(4+5\cos \theta )}^2+9{\sin }^2\theta }{{(5+4\cos \theta )}^2} \\ =\frac{(16+40\cos \theta +25{\cos }^2\theta +9{\sin }^2\theta )}{{(5+4\cos \theta )}^2} \\ =\frac{16+40\cos \theta +16{\cos }^2\theta +9({\sin }^2\theta +{\cos }^2\theta )}{{(5+4\cos \theta )}^2} \\ =\frac{16+40\cos \theta +16{\cos }^2\theta +9}{{(5+4\cos \theta )}^2} \\ =\frac{16{\cos }^2\theta +40\cos \theta +25}{{(5+4\cos \theta )}^2} \\ =\frac{{(4\cos \theta +5)}^2}{{(5+4\cos \theta )}^2} \\ =\mathbf{1} \end{gathered}$$

Hence, Option ‘$B$’ is Correct.