The real value of theta for which the express | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $z = \frac{1 + i \cos 	heta}{1 - 2i \cos 	heta}$.To make $z$ a real number,we multiply the numerator and denominator by the conjugate of the

Vedclass · Accepted Answer

$(2n + 1)\frac{\pi}{2}$

Vedclass · Answer

(B) Let $z = \frac{1 + i \cos 	heta}{1 - 2i \cos 	heta}$.To make $z$ a real number,we multiply the numerator and denominator by the conjugate of the denominator,which is $1 + 2i \cos 	heta$:$z = \frac{(1 + i \cos 	heta)(1 + 2i \cos 	heta)}{(1 - 2i \cos 	heta)(1 + 2i \cos 	heta)}$$z = \frac{1 + 2i \cos 	heta + i \cos 	heta + 2i^2 \cos^2 	heta}{1 + 4 \cos^2 	heta}$Since $i^2 = -1$,we have:$z = \frac{1 - 2 \cos^2 	heta + 3i \cos 	heta}{1 + 4 \cos^2 	heta}$For $z$ to be a real number,the imaginary part must be zero:$\frac{3 \cos 	heta}{1 + 4 \cos^2 	heta} = 0$This implies $\cos 	heta = 0$.The general solution for $\cos 	heta = 0$ is $	heta = (2n + 1)\frac{\pi}{2}$ for $n \in I$.Thus,the correct option is $(B)$.

The real value of $\theta$ for which the expression $\frac{1 + i \cos \theta}{1 - 2i \cos \theta}$ is a real number is $(n \in I)$:

Similar Questions

Let complex number $z$ be such that $|z - \frac{6}{z}| = 5$,then the maximum value of $|z|$ will be -

If complex numbers $(x - 2y) + i(3x - y)$ and $(2x - y) + i(x - y + 6)$ are conjugates of each other,then $|x + iy|$ is $(x, y \in \mathbb{R})$.

The Cartesian form of the complex number $z = 4(\cos 300^{\circ} + i \sin 300^{\circ})$ is

If $x + \frac{1}{x} = \sqrt{3},$ then $x =$

The product of two complex numbers,each of unit modulus,is also a complex number of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module