Let Z and W be complex numbers such that |Z| | vedclass

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

Question

(B) Let $|Z| = |W| = r$. Then $Z = r e^{i	heta}$ and $W = r e^{i\phi}$,where $	heta = 	ext{arg } Z$ and $\phi = 	ext{arg } W$. Given $	heta + \ph

Vedclass · Accepted Answer

Statement $1$ is true,Statement $2$ is true,Statement $2$ is a correct explanation for Statement $1$.

Vedclass · Answer

(B) Let $|Z| = |W| = r$. Then $Z = r e^{i	heta}$ and $W = r e^{i\phi}$,where $	heta = 	ext{arg } Z$ and $\phi = 	ext{arg } W$. Given $	heta + \phi = \pi$,we have $	heta = \pi - \phi$. Thus,$Z = r e^{i(\pi - \phi)} = r e^{i\pi} e^{-i\phi} = -r e^{-i\phi}$. Since $\overline{W} = r e^{-i\phi}$,we get $Z = -\overline{W}$. So,Statement $1$ is true. For Statement $2$,$	ext{arg } \overline{W} = -	ext{arg } W = -\phi$. Then $	ext{arg } Z - 	ext{arg } \overline{W} = 	heta - (-\phi) = 	heta + \phi = \pi$. Thus,Statement $2$ is also true and explains Statement $1$.

Let $Z$ and $W$ be complex numbers such that $|Z| = |W|$,and $\text{arg } Z$ denotes the principal argument of $Z$.
Statement $1$: If $\text{arg } Z + \text{arg } W = \pi$,then $Z = -\overline{W}$.
Statement $2$: $|Z| = |W|$ implies $\text{arg } Z - \text{arg } \overline{W} = \pi$.

Similar Questions

$z = \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}, \quad (i = \sqrt{-1})$ will be purely imaginary if $\theta =$

If $x$ and $y$ are two positive real numbers such that $x+iy = \frac{13 \sqrt{-5+12i}}{(2-3i)(3+2i)}$,then $13y-26x=$

The number of all possible solutions of the equation $z^3+\overline{z}=0$ is

$\frac{1 + 7i}{(2 - i)^2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $Z$ and $W$ be complex numbers such that $|Z| = |W|$,and $\text{arg } Z$ denotes the principal argument of $Z$.Statement $1$: If $\text{arg } Z + \text{arg } W = \pi$,then $Z = -\overline{W}$.Statement $2$: $|Z| = |W|$ implies $\text{arg } Z - \text{arg } \overline{W} = \pi$.