Question

Find square root of $9+40i$

Answer 1

Let $x+iy=\sqrt{9+40i}$
$(x-iy)=9+40i$
$\therefore$ $x^2+y^2=9$ (i)
and $xy=20$ (ii)
squaring (i) and adding with $4$ times the square of (ii)
we get $x^4+y^4-2x^2{y}^2+4x^2{y}^2=81+1600$
$\Rightarrow$ ${(x^2+y^2)}^2=1681$
$\Rightarrow$ $x^2+y^2=41$ (iii)
from (i)+(iii) we get $x^2=25$
$\Rightarrow$ $x=\pm 5$
and $y^2=16$
$\Rightarrow$ $y=\pm 4$
from equation (ii) we can see that
$x$ & $y$ are of same sign
$\therefore$ $x+iy=(5+4i)$ or $-(5+4i)$
$\therefore$ sq. roots of $9+40i=\pm (5+4i)$
$\pm (5+4i)$