If x, y in R and x^2+y+4 i and -3+x^2 y i are | vedclass

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

Question

(C) Given that $x^2+y+4 i$ and $-3+x^2 y i$ are conjugates to each other.Therefore,$x^2+y+4 i = -3 - x^2 y i$.Comparing the real and imaginary parts o

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) Given that $x^2+y+4 i$ and $-3+x^2 y i$ are conjugates to each other.Therefore,$x^2+y+4 i = -3 - x^2 y i$.Comparing the real and imaginary parts on both sides:$x^2+y = -3$  $(i)$$4 = -x^2 y \Rightarrow y = -\frac{4}{x^2}$  $(ii)$Substituting $(ii)$ into $(i)$:$x^2 - \frac{4}{x^2} = -3$$x^4 - 4 = -3x^2$$x^4 + 3x^2 - 4 = 0$$(x^2+4)(x^2-1) = 0$Since $x \in R$,$x^2$ must be non-negative,so $x^2+4 
eq 0$.Thus,$x^2-1 = 0$ $\Rightarrow x^2 = 1$ $\Rightarrow x = \pm 1$.Substituting $x^2=1$ into $(ii)$:$y = -\frac{4}{1} = -4$.Now,$(|x|+|y|)^2 = (|\pm 1| + |-4|)^2 = (1+4)^2 = 5^2 = 25$.

If $x, y \in R$ and $x^2+y+4 i$ and $-3+x^2 y i$ are conjugates to each other,then $(|x|+|y|)^2=$

Similar Questions

For any complex number $z$,$\bar{z} = \frac{1}{z}$ if and only if

If $a > 0$ and $z = \frac{(1 + i)^2}{a - i}$,where $i = \sqrt{-1}$,has magnitude $\sqrt{\frac{2}{5}}$,then $\overline{z}$ is equal to

The number of integer solutions of the equation $|1-i|^x=2^x$ is

If $\alpha$ and $\beta$ are different complex numbers with $|\beta| = 1$,then $\left| \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right|$ is equal to

Let $\alpha$ be a fixed non-zero complex number with $|\alpha| < 1$ and $w = \frac{z-\alpha}{1-\bar{\alpha}z}$,where $z$ is a complex number. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module