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

If $a = \cos \theta + i \sin \theta$,then $\frac{1 + a}{1 - a} = $

For any two complex numbers $z_1$ and $z_2$ and any real numbers $a$ and $b$,what is the value of $|az_1 - bz_2|^2 + |bz_1 + az_2|^2$?

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

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

Find the multiplicative inverse of the complex number $4-3i$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module