Finding the squareroot of given complex number

Exercise:

Find the Square Root of $z=8+i6$

Solution: Let $w\in \mathbb{C}$ be the square root of $z$ hence $w=\sqrt{z}\Leftrightarrow {{w}^{2}}=z$

Let $w=x+iy\Leftrightarrow {{w}^{2}}={{x}^{2}}-{{y}^{2}}+i2xy=8+i6$

But ${{\left| w \right|}^{2}}=\left| z \right|\Leftrightarrow {{x}^{2}}+{{y}^{2}}=10$ so ${{x}^{2}}-{{y}^{2}}=8\,\,\,,\,\,2xy=6\,$

Hence $2{{x}^{2}}=18\Leftrightarrow {{x}^{2}}=9\Leftrightarrow x=\pm 3$ and $-2{{y}^{2}}=8-10=-2\Leftrightarrow {{y}^{2}}=1\Leftrightarrow y=\pm 1$

Since $xy=3>0$ hence $x\,\And y$ must have same signs

Thus ${{w}_{1}}=3+i\,\,\And \,\,{{w}_{2}}=-3-i$