Question

Find the roots of the complex number $z=4-3i$

Answer 1

Given,

$z=4-3i$

$(x+yi)(x+yi)=4-3i$

$x^2-y^2+2ixy=4-3i$

$x^2-y^2=4$.....(1)

$2xy=-3$

$xy=-\frac{3}{2}$

$x=-\frac{3}{2y}$

From (1)

${(-\frac{3}{2y})}^2-y^2=4$

$\frac{9}{4y^2}-y^2=4$

$4y^4+16y^2-9=0$

$(2y^2-1)(2y^2+9)=0$

$y^2=\frac{1}{2}$

$\therefore y=\pm \frac{1}{\sqrt{2}}$

when, $y=\frac{1}{\sqrt{2}},x=-\frac{3}{\sqrt{2}}$

when, $y=-\frac{1}{\sqrt{2}},x=\frac{3}{\sqrt{2}}$

$\therefore \sqrt{(-\frac{3}{\sqrt{2}}+i\frac{1}{\sqrt{2}})}=4-3i$

$\therefore \sqrt{(\frac{3}{\sqrt{2}}-i\frac{1}{\sqrt{2}})}=4-3i$