Question:
The values of \(\theta\), for which
\[
\frac{3+2i\sin\theta}{1-2i\sin\theta}
\]
is real are
The values of \(\theta\), for which
\[
\frac{3+2i\sin\theta}{1-2i\sin\theta}
\]
is real are
Show Hint
To check when a complex expression is real, rationalize the denominator and make the imaginary part equal to zero.
Updated On: Jun 26, 2026
- \(\theta=n\pi+\dfrac{\pi}{3}\) for \(n\in \mathbb{Z}\)
- \(\theta=n\pi+\dfrac{\pi}{4}\) for \(n\in \mathbb{Z}\)
- \(\theta=n\pi+\dfrac{\pi}{2}\) for \(n\in \mathbb{Z}\)
- \(\theta=n\pi\) for \(n\in \mathbb{Z}\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Let \(\sin\theta=s\).
Given expression is \[ \frac{3+2i\sin\theta}{1-2i\sin\theta} \] Let \[ s=\sin\theta \] Then the expression becomes \[ \frac{3+2is}{1-2is} \]
Step 2: Rationalize the denominator.
Multiply numerator and denominator by the conjugate of the denominator: \[ \frac{3+2is}{1-2is}\times \frac{1+2is}{1+2is} \] \[ =\frac{(3+2is)(1+2is)}{(1-2is)(1+2is)} \]
Step 3: Simplify the numerator and denominator.
Numerator: \[ (3+2is)(1+2is) \] \[ =3+6is+2is+4i^2s^2 \] Since \[ i^2=-1, \] we get \[ 3+8is-4s^2 \] So the numerator is \[ 3-4s^2+8is \] Denominator: \[ (1-2is)(1+2is)=1+4s^2 \] Therefore, \[ \frac{3+2is}{1-2is} = \frac{3-4s^2+8is}{1+4s^2} \]
Step 4: Apply the condition for the expression to be real.
For the expression to be real, its imaginary part must be zero.
The imaginary part is \[ \frac{8s}{1+4s^2} \] So, \[ \frac{8s}{1+4s^2}=0 \] Since \[ 1+4s^2\neq 0, \] we must have \[ s=0 \] Thus, \[ \sin\theta=0 \]
Step 5: Find the values of \(\theta\).
We know that \[ \sin\theta=0 \] when \[ \theta=n\pi,\quad n\in \mathbb{Z} \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\theta=n\pi,\quad n\in \mathbb{Z}} \]
Given expression is \[ \frac{3+2i\sin\theta}{1-2i\sin\theta} \] Let \[ s=\sin\theta \] Then the expression becomes \[ \frac{3+2is}{1-2is} \]
Step 2: Rationalize the denominator.
Multiply numerator and denominator by the conjugate of the denominator: \[ \frac{3+2is}{1-2is}\times \frac{1+2is}{1+2is} \] \[ =\frac{(3+2is)(1+2is)}{(1-2is)(1+2is)} \]
Step 3: Simplify the numerator and denominator.
Numerator: \[ (3+2is)(1+2is) \] \[ =3+6is+2is+4i^2s^2 \] Since \[ i^2=-1, \] we get \[ 3+8is-4s^2 \] So the numerator is \[ 3-4s^2+8is \] Denominator: \[ (1-2is)(1+2is)=1+4s^2 \] Therefore, \[ \frac{3+2is}{1-2is} = \frac{3-4s^2+8is}{1+4s^2} \]
Step 4: Apply the condition for the expression to be real.
For the expression to be real, its imaginary part must be zero.
The imaginary part is \[ \frac{8s}{1+4s^2} \] So, \[ \frac{8s}{1+4s^2}=0 \] Since \[ 1+4s^2\neq 0, \] we must have \[ s=0 \] Thus, \[ \sin\theta=0 \]
Step 5: Find the values of \(\theta\).
We know that \[ \sin\theta=0 \] when \[ \theta=n\pi,\quad n\in \mathbb{Z} \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\theta=n\pi,\quad n\in \mathbb{Z}} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Complex numbers Questions
- If \( z \) is a complex number such that \( |z| = 5 \) and \( \text{Re}(z) = 3 \), then the value of \( z^2 \) is:
- \( \frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y+i}{3-i} = i \implies x+y = \)
- Let z and w be two distinct non-zero complex numbers if \(|z|^2w - |w|^2z = z - w\), then
- If 1, \(\omega, \omega^2\) denote the cube roots of unity, then, the value of \( (1-\omega+\omega^2)^5 + (1+\omega-\omega^2)^5 \) is
- The values of \(\theta\), for which \( \frac{3+2i\sin\theta}{1-2i\sin\theta} \) is real are
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions