If z = r{e^{itheta }}, then |{e^{iz}}| = | vedclass

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

Question

(B) Given $z = r{e^{i	heta }} = r(\cos 	heta + i\sin 	heta )$.Then $iz = ir(\cos 	heta + i\sin 	heta ) = -r\sin 	heta + ir\cos 	heta$.Therefore

Vedclass · Accepted Answer

${e^{ - r\sin 	heta }}$

Vedclass · Answer

(B) Given $z = r{e^{i	heta }} = r(\cos 	heta + i\sin 	heta )$.Then $iz = ir(\cos 	heta + i\sin 	heta ) = -r\sin 	heta + ir\cos 	heta$.Therefore,${e^{iz}} = {e^{-r\sin 	heta + ir\cos 	heta}} = {e^{-r\sin 	heta}} \cdot {e^{i(r\cos 	heta)}}$.Taking the modulus on both sides:$|{e^{iz}}| = |{e^{-r\sin 	heta}}| \cdot |{e^{i(r\cos 	heta)}}|$.Since ${e^{-r\sin 	heta}}$ is a real number,$|{e^{-r\sin 	heta}}| = {e^{-r\sin 	heta}}$.Also,$|{e^{i(r\cos 	heta)}}| = |\cos(r\cos 	heta) + i\sin(r\cos 	heta)| = \sqrt{\cos^2(r\cos 	heta) + \sin^2(r\cos 	heta)} = 1$.Thus,$|{e^{iz}}| = {e^{-r\sin 	heta}} \cdot 1 = {e^{-r\sin 	heta}}$.

If $z = r{e^{i\theta }},$ then $|{e^{iz}}| = $

Similar Questions

Let $z_{1} = 2 - i$ and $z_{2} = -2 + i$. Find $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right)$.

$\left|\frac{1}{i^{2020}}+\frac{2}{i^{2021}}+\frac{3}{i^{2022}}+\frac{4}{i^{2023}}\right|$ is equal to

If $a = \cos \theta + i \sin \theta$,then $\frac{1 + a}{1 - a} = $

Let $\alpha$ be a fixed non-zero complex number with $|\alpha| < 1$ and $w = \frac{z-\alpha}{1-\bar{\alpha}z}$,where $z$ is a complex number. Then,

If $z$ is a complex number,then $(\overline {{z^{ - 1}}} )(\overline z ) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module