# If (x+iy)^3 is an imaginary number and |x+iy|=8, find x and y?

Jul 4, 2017

$x = \pm 4 \sqrt{3} \mathmr{and} y = \pm 4$

#### Explanation:

${\left(x + i y\right)}^{3}$
$= {x}^{3} + 3 {x}^{2} i y + 3 x {\left(i y\right)}^{2} + {\left(i y\right)}^{3}$
$= {x}^{3} + 3 {x}^{2} i y - 3 x {y}^{2} - i {y}^{3}$
$\left({x}^{3} - 3 x {y}^{2}\right) + i \left(3 {x}^{2} y - {y}^{3}\right)$
As the value is purely imaginary hence
$\left({x}^{3} - 3 x {y}^{2}\right) = 0$
${x}^{2} = 3 {y}^{2}$
It's also given
$x + i y = 8$
=>।x+iy।=8
$\sqrt{{x}^{2} + {y}^{2}} = 8$

i.e. ${x}^{2} + {y}^{2} = 64$ or $3 {y}^{2} + {y}^{2} = 64$

or $4 {y}^{2} = 64$ i.e. $y = \pm 4$

and ${x}^{2} = 48$ i.e. $x = \pm 4 \sqrt{3}$
From those equations we get the answer.