Let Z_1 and Z_2 be any two complex numbers.St | vedclass

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

Question

(B) Statement $1$ is the triangle inequality for subtraction,which states $|Z_1 - Z_2| \ge ||Z_1| - |Z_2||$,and since $||Z_1| - |Z_2|| \ge |Z_1| - |Z_

Vedclass · Accepted Answer

Statement $1$ is true,Statement $2$ is true,Statement $2$ is not a correct explanation of Statement $1$.

Vedclass · Answer

(B) Statement $1$ is the triangle inequality for subtraction,which states $|Z_1 - Z_2| \ge ||Z_1| - |Z_2||$,and since $||Z_1| - |Z_2|| \ge |Z_1| - |Z_2|$,Statement $1$ is true.Statement $2$ is the standard triangle inequality for addition,which is also a fundamental property of complex numbers and is true.However,Statement $2$ does not explain Statement $1$ as they are independent properties of the modulus of complex numbers.

Let $Z_1$ and $Z_2$ be any two complex numbers.
Statement $1: |Z_1 - Z_2| \ge |Z_1| - |Z_2|$
Statement $2: |Z_1 + Z_2| \le |Z_1| + |Z_2|$

Similar Questions

If $m_1, m_2, m_3$ and $m_4$ respectively denote the moduli of the complex numbers $1+4 i, 3+i, 1-i$ and $2-3 i$,then the correct one,among the following is

If $a=|\bar{a}|$ and $b=|\bar{b}|$,then $\left(\frac{\bar{a}}{a^2}-\frac{\bar{b}}{b^2}\right)^2=$

Sum of the moduli of the complex roots of the equation $(x^2+\frac{1}{x^2})-5(x+\frac{1}{x})+6=0$ is

If $x$ and $y$ are real numbers,which of the following is always true?

Let $z = a - \frac{i}{2}$,where $a \in R$. Then $|i + z|^2 - |i - z|^2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $Z_1$ and $Z_2$ be any two complex numbers.Statement $1: |Z_1 - Z_2| \ge |Z_1| - |Z_2|$Statement $2: |Z_1 + Z_2| \le |Z_1| + |Z_2|$