# Find real values of x and y for which

Question:

Find real values of x and y for which

$\left(x^{4}+2 x i\right)-\left(3 x^{2}+i y\right)=(3-5 i)+(1+2 i y)$

Solution:

We have, $\left(x^{4}+2 x i\right)-\left(3 x^{2}+i y\right)=(3-5 i)+(1+2 i y)$

$\Rightarrow x^{4}+2 x i-3 x^{2}+i y=3-5 i+1+2 i y$

$\Rightarrow\left(x^{4}-3 x^{2}\right)+i(2 x-y)=4+i(2 y-5)$

On equating real and imaginary parts, we get

$x^{4}-3 x^{2}=4$ and $2 x-y=2 y-5$

$\Rightarrow x^{4}-3 x^{2}-4=0$ eq(i) and $2 x-y-2 y+5=0$ eq(ii)

Now from eq (i), $x^{4}-3 x^{2}-4=0$

$\Rightarrow x^{4}-4 x^{2+} x^{2}-4=0$

$\Rightarrow x^{2}\left(x^{2}-4\right)+1\left(x^{2}-4\right)=0$

$\Rightarrow\left(x^{2}-4\right)\left(x^{2}+1\right)=0$

$\Rightarrow x^{2}-4=0$ and $x^{2}+1=0$

$\Rightarrow x=\pm 2$ and $x=\sqrt{-1}$

Real value of $x=\pm 2$

Putting $x=2$ in eq (ii), we get

$2 x-3 y+5=0$

$\Rightarrow 2 \times 2-3 y+5=0$

$\Rightarrow 4-3 y+5=0=9-3 y=0$

$\Rightarrow y=3$

Putting $x=-2$ in eq (ii), we get

$2 x-3 y+5=0$

$\Rightarrow 2 x-2-3 y+5=0$

$\Rightarrow-4-3 y+5=0=1-3 y=0$

$\Rightarrow y=\frac{1}{3}$