Question

$(x^4+2xi)-(3x^2+yi)=(1+2yi)$. Find x and y .

Answer 1

$(x^4+2xi)-(3x^2+yi)=(1+2yi)$

$(x^4-3x^2)+(2x-y)i=1+2yi$

As we know that two complex numbers are equal if their corresponding real and imaginary parts are equal.

Therefore,

$x^4-3x^2=1.....\left(1\right)$

$2x-y=2y$

$y=\frac{2}{3}x.....\left(2\right)$

Solving ${eq}^n\left(1\right)$, we have

$x^4-3x^2-1=0$

${\left(x^2\right)}^2-3x^2-1=0$

By quadratic formula,

$x^2=\frac{-(-3)\pm \sqrt{{(-3)}^2-4\times (-1)\times 1}}{2\times 1}$

$\Rightarrow x^2=\frac{3\pm \sqrt{9+4}}{2}$

$\Rightarrow x=\pm \sqrt{\frac{3\pm \sqrt{13}}{2}}$

Substituting the value of $x$ in ${eq}^n\left(2\right)$, we have

$y=\frac{2}{3}\times (\pm \sqrt{\frac{3\pm \sqrt{13}}{2}})$

$\Rightarrow y=\frac{1}{3}(\pm \sqrt{2(3\pm \sqrt{13})})$

$\Rightarrow y=\frac{1}{3}(\pm \sqrt{6\pm 2\sqrt{13}})$

Thus the value of $x$ and $y$ are $(\pm \sqrt{\frac{3\pm \sqrt{13}}{2}})$ and $\frac{1}{3}(\pm \sqrt{6\pm 2\sqrt{13}})$ respectively.