User:Jiujiu1123/sandbox

Let z = X + iY

$$z = \sqrt{12+5i}$$

$$z^2 = 12 + 5i$$

$$(X+iY)^2 = 12+5i$$

$$X^2 +2XYi - Y^2 = 12+5i$$

Equating Real Part

$$X^2-Y^2 = 12$$(equation 1)

Equating Imaginary Part

$$2XY = 5$$(equation 2)

$$X = \frac{5}{2Y}$$

Plug back in to Equation 1.

$$(\frac{5}{2Y})^2-Y^2 = 12$$

$$\frac{25}{4Y^2}-Y^2 = 12$$

$$25-4Y^4 = 48Y^2$$

Let $$Y^2 = u$$

$$25-4u^2 = 48u$$

$$4u^2+48u-25 =0$$

As $$Y \in \mathbb{R}$$, then $$Y^2\geq 0 $$, then $$u\geq 0 $$

$$u = \frac{-48+\sqrt{48^2-4*4*-25}}{2*4}$$

$$u = \frac{-48+52}{8}$$

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

$$Y = \pm\frac{1}{\sqrt{2}}$$

When $$Y = \frac{1}{\sqrt{2}}$$

$$\frac{2}{\sqrt{2}}X=5$$

$$X = \frac{5\sqrt{2}}{2}$$

When $$Y = -\frac{1}{\sqrt{2}}$$

$$-\frac{2}{\sqrt{2}}X=5$$

$$X = -\frac{5\sqrt{2}}{2}$$

Therefore $$z = \pm(\frac{5\sqrt{2}}{2}}+{\frac{1}{\sqrt{2}}i)$$