# Find the square root of 4-3i?

Aug 7, 2018

$z = \pm \frac{3 - i}{2} \sqrt{2}$

#### Explanation:

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

${x}^{2} - {y}^{2} = 4$
$2 x y = - 3 \implies y = - \frac{3}{2 x}$

${x}^{2} - \frac{9}{4 {x}^{2}} = 4 , {x}^{2} = a$

$4 {a}^{2} - 16 a - 9 = 0$

$\Delta = 256 + 4 \cdot 4 \cdot 9 = 400$

${x}^{2} = \frac{16 \pm 20}{8} > 0$

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

$y = - \pm \frac{3}{2} \cdot \frac{\sqrt{2}}{3} = - \pm \frac{1}{2} \sqrt{2}$

Aug 8, 2018

$4 - 3 i$

$= \frac{1}{2} \left(8 - 2.3 \cdot i\right)$

$= \frac{1}{2} \left({3}^{2} + {i}^{2} - 2.3 \cdot i\right)$

$= \frac{1}{2} {\left(3 - i\right)}^{2}$

Square root of $4 - 3 i$

$\pm \frac{1}{\sqrt{2}} \left(3 - i\right)$