The amplitude of {e^{{e^{ - itheta }}}} is eq | vedclass

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

Question

(B) Let $z = {e^{{e^{ - i	heta }}}} = {e^{\cos 	heta - i\sin 	heta }} = {e^{\cos 	heta }}{e^{ - i\sin 	heta }}$.$z = {e^{\cos 	heta }}[\cos (\si

Vedclass · Accepted Answer

$ - \sin 	heta $

Vedclass · Answer

(B) Let $z = {e^{{e^{ - i	heta }}}} = {e^{\cos 	heta - i\sin 	heta }} = {e^{\cos 	heta }}{e^{ - i\sin 	heta }}$.$z = {e^{\cos 	heta }}[\cos (\sin 	heta ) - i\sin (\sin 	heta )]$.$z = {e^{\cos 	heta }}\cos (\sin 	heta ) - i{e^{\cos 	heta }}\sin (\sin 	heta )$.$amp(z) = {	an ^{ - 1}}\left[ { - \frac{{{e^{\cos 	heta }}\sin (\sin 	heta )}}{{{e^{\cos 	heta }}\cos (\sin 	heta )}}} ight]$.$amp(z) = {	an ^{ - 1}}[	an ( - \sin 	heta )] = - \sin 	heta $.

The amplitude of ${e^{{e^{ - i\theta }}}}$ is equal to

Similar Questions

If $-1 + \sqrt{-3} = re^{i\theta}$, then the value of $\theta$ is

If $i=\sqrt{-1}$,then $\operatorname{Arg}\left[\frac{(1+i)^{2025}}{(1-i)^{2022}}\right]=$

If $z = \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}$,then $arg(z) = $ ............. $^\circ$

Let $z_1$ and $z_2$ be two non-zero complex numbers. Then

Let $z_1, z_2$ be two complex numbers such that $\bar{z}_1 - i \bar{z}_2 = 0$ and $\arg(z_1 z_2) = \frac{3 \pi}{4}$,then $\arg(z_1) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module