If $x, y \in R$ and $x^2+y+4 i$ and $-3+x^2 y i$ are conjugates to each other, then $(|x|+|y|)^2=$
If $x, y \in R$ and $x^2+y+4 i$ and $-3+x^2 y i$ are conjugates to each other, then $(|x|+|y|)^2=$
17
16
25
9
Solution
$x^2+y+4 i$ and $-3+x^2 y i$ are conjugate.
Therefore, $x^2+y+4 i=-3-x^2 y i$
$
\Rightarrow \quad\left(x^2+y\right)+4 i=(-3)-x^2 y i
$
On comparing both sides, we get
$
\begin{array}{ll}
\Rightarrow & x^2+y=-3 \\
\text { and } & 4=-x^2 y \\
\Rightarrow & y=\frac{4}{-x^2}=-\frac{4}{x^2}
\end{array}
$
On puting the value of $y$ in Eq. (i), we get
$
\begin{array}{ll}
\therefore & x^2-\frac{4}{x^2}=-3 \Rightarrow x^4-4=-3 x^2 \\
\Rightarrow & x^4+3 x^2-4=0 \\
\Rightarrow \quad & \left(x^2+4\right)\left(x^2-1\right)=0 \\
\Rightarrow \quad & x^2+4=0 \\
\Rightarrow \quad & x^2=-4 \\
& x^2-1=0 \\
\Rightarrow \quad & x^2=1 \Rightarrow x= \pm 1
\end{array}
$
Put the value of $x= \pm 1$ in Eq. (ii), we get
$
\begin{aligned}
& \qquad \begin{array}{l}
y=\frac{-4}{(1)^2}=-4 \\
x \\
\text { Hence, }\left(|x|+\left.|y|\right|^2\right. \\
=|x|^2+|y|^2+2|x||y| \\
=(1)^2+|(-4)|^2+2|1||-4| \\
=1+16+8=25
\end{array}
\end{aligned}
$