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

For real numbers $a$ and $b$,if $4a + i(3a - b) = b - 6i$ and $z = a + \frac{b}{4}i$,then $\frac{|z|}{a} = $

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 $z$ is a complex number such that $\left|z-\frac{4}{z}\right|=2$,then the greatest value of $|z|$ is

$(z + a)(\bar z + a)$,where $a$ is real,is equivalent to

If $z_1 = 1+2i$ and $z_2 = 3+5i$,then $\text{Re} \left( \frac{\bar{z}_2 z_1}{z_2} \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module