Question:
If
\[
\frac{3}{2+\cos\theta+i\sin\theta}=x+iy,
\]
then
\[
(x-1)(x-3)=
\]
If
\[
\frac{3}{2+\cos\theta+i\sin\theta}=x+iy,
\]
then
\[
(x-1)(x-3)=
\]
Show Hint
For expressions involving $x+iy$, rationalize the denominator first and compare real and imaginary parts.
Updated On: Jun 3, 2026
- $y^2$
- $-y^2$
- $0$
- $1$
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Express the complex number in the form $x+iy$ by rationalizing the denominator.
Step 2: Meaning
Let \[ z=\frac{3}{2+\cos\theta+i\sin\theta}. \] Multiply numerator and denominator by the conjugate \[ 2+\cos\theta-i\sin\theta. \]
Step 3: Analysis
Then \[ x+iy= \frac{3(2+\cos\theta-i\sin\theta)} {(2+\cos\theta)^2+\sin^2\theta}. \] Since \[ (2+\cos\theta)^2+\sin^2\theta =5+4\cos\theta, \] we get \[ x=\frac{3(2+\cos\theta)}{5+4\cos\theta}, \qquad y=-\frac{3\sin\theta}{5+4\cos\theta}. \] Now, \[ x-1=\frac{-2-\cos\theta}{5+4\cos\theta}, \] and \[ x-3=\frac{-9(1+\cos\theta)}{5+4\cos\theta}. \] Therefore, \[ (x-1)(x-3) = -\frac{9\sin^2\theta}{(5+4\cos\theta)^2}. \]
Step 4: Conclusion
Since \[ y^2=\frac{9\sin^2\theta}{(5+4\cos\theta)^2}, \] it follows that \[ (x-1)(x-3)=-y^2. \]
Final Answer: (B)
Express the complex number in the form $x+iy$ by rationalizing the denominator.
Step 2: Meaning
Let \[ z=\frac{3}{2+\cos\theta+i\sin\theta}. \] Multiply numerator and denominator by the conjugate \[ 2+\cos\theta-i\sin\theta. \]
Step 3: Analysis
Then \[ x+iy= \frac{3(2+\cos\theta-i\sin\theta)} {(2+\cos\theta)^2+\sin^2\theta}. \] Since \[ (2+\cos\theta)^2+\sin^2\theta =5+4\cos\theta, \] we get \[ x=\frac{3(2+\cos\theta)}{5+4\cos\theta}, \qquad y=-\frac{3\sin\theta}{5+4\cos\theta}. \] Now, \[ x-1=\frac{-2-\cos\theta}{5+4\cos\theta}, \] and \[ x-3=\frac{-9(1+\cos\theta)}{5+4\cos\theta}. \] Therefore, \[ (x-1)(x-3) = -\frac{9\sin^2\theta}{(5+4\cos\theta)^2}. \]
Step 4: Conclusion
Since \[ y^2=\frac{9\sin^2\theta}{(5+4\cos\theta)^2}, \] it follows that \[ (x-1)(x-3)=-y^2. \]
Final Answer: (B)
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