left| {(1 + i)frac{{(2 + i)}}{{(3 + i)}}} rig | vedclass

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

Question

(C) We use the property of the modulus of complex numbers: $|z_1 z_2| = |z_1| |z_2|$ and $|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$.$\left| {(1 + i)\fr

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) We use the property of the modulus of complex numbers: $|z_1 z_2| = |z_1| |z_2|$ and $|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$.$\left| {(1 + i)\frac{{(2 + i)}}{{(3 + i)}}} ight| = |1 + i| 	imes \frac{|2 + i|}{|3 + i|}$$= \sqrt{1^2 + 1^2} 	imes \frac{\sqrt{2^2 + 1^2}}{\sqrt{3^2 + 1^2}}$$= \sqrt{2} 	imes \frac{\sqrt{5}}{\sqrt{10}}$$= \frac{\sqrt{10}}{\sqrt{10}} = 1$.

$\left| {(1 + i)\frac{{(2 + i)}}{{(3 + i)}}} \right| = $

Similar Questions

Let ${z_1}$ be a complex number with $|{z_1}| = 1$ and ${z_2}$ be any complex number,then $\left| \frac{{z_1 - z_2}}{{1 - z_1 \bar{z}_2}} \right| = $

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

Find the multiplicative inverse of the complex number $\sqrt{5}+3i$.

If $z = \frac{4}{1-i}$,then $\bar{z}$ is (where $\bar{z}$ is the complex conjugate of $z$).

Find the conjugate of $\frac{5i}{7+i}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module