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

The conjugate of $\frac{(2 + i)^2}{3 + i}$ in the form of $a + ib$ is

The real part of $(1 - \cos \theta + 2i\sin \theta )^{-1}$ is

If $Z_1=2+i$ and $Z_2=3-4i$,and $\frac{\overline{Z_1}}{\overline{Z_2}}=a+bi$,then the value of $-7a+b$ is (where $i=\sqrt{-1}$ and $a, b \in \mathbb{R}$)

Let $Z$ be a complex number such that $|Z|+Z=2+i$ (where $i=\sqrt{-1}$),then $|Z|$ is equal to

The conjugate of the complex number $\frac{2 - 3i}{4 - i}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module