The real part of frac{1}{1 - cos theta + i si | vedclass

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

Question

(B) Given expression: $z = \frac{1}{(1 - \cos 	heta) + i \sin 	heta}$.Multiply the numerator and denominator by the conjugate $(1 - \cos 	heta) - i

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(B) Given expression: $z = \frac{1}{(1 - \cos 	heta) + i \sin 	heta}$.Multiply the numerator and denominator by the conjugate $(1 - \cos 	heta) - i \sin 	heta$:$z = \frac{(1 - \cos 	heta) - i \sin 	heta}{(1 - \cos 	heta)^2 + \sin^2 	heta}$.Simplify the denominator:$(1 - \cos 	heta)^2 + \sin^2 	heta = 1 - 2 \cos 	heta + \cos^2 	heta + \sin^2 	heta = 1 - 2 \cos 	heta + 1 = 2 - 2 \cos 	heta = 2(1 - \cos 	heta)$.Thus,$z = \frac{1 - \cos 	heta}{2(1 - \cos 	heta)} - i \frac{\sin 	heta}{2(1 - \cos 	heta)}$.The real part is $\frac{1 - \cos 	heta}{2(1 - \cos 	heta)} = \frac{1}{2}$.

The real part of $\frac{1}{1 - \cos \theta + i \sin \theta}$ is equal to

Similar Questions

If $z = 1 + \cos \theta - i \sin \theta$ and $0 < \theta < \pi$,then $\left[|z - 1|^2 - \frac{|z|^2}{4}\right]^{1/2} =$

If $m_1, m_2, m_3$ and $m_4$ respectively denote the moduli of the complex numbers $1+4 i, 3+i, 1-i$ and $2-3 i$,then the correct one,among the following is

The conjugate of $1 + i$ is

If $x+iy=\sqrt{\frac{3+i}{1+3i}}$,then $\left(x^2+y^2\right)^2=$

If ${z_1}$ and ${z_2}$ are two complex numbers,then $|{z_1} + {z_2}|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module