Solving equation of 2nd degree with variable is complex number SG-Level

Exercise:

Solve in $\mathbb{C},$ ${{z}^{2}}+\left( 1+i \right)z+2i=0$

Solution: First let’s find the discernment $\Delta ={{b}^{2}}-4ac={{\left( 1+i \right)}^{2}}-4\left( 2i \right)=1-1+2i-8i=-6i$

So there is $\omega \in \mathbb{C}$ such that $\Delta ={{\omega }^{2}}$ where $w=x+iy\Leftrightarrow {{\omega }^{2}}={{x}^{2}}-{{y}^{2}}+i2xy$

So ${{x}^{2}}-{{y}^{2}}=0\,\,,\,\,2xy=-6\,\,\,\,\And \,\,{{x}^{2}}+{{y}^{2}}=6\,$

So $2{{x}^{2}}=6\Leftrightarrow {{x}^{2}}=3\Leftrightarrow x=\pm \sqrt{3}$  and $y=\pm \sqrt{3}$

Since $xy=-3<0$ then $x\,\,\And \,\,y$ must have different signs

Hence ${{w}_{1}}=\sqrt{3}\left( -1+i \right)\,\,\,\And \,\,{{w}_{2}}=\sqrt{3}\left( 1-i \right)$

So the roots are :

${{z}_{1}}=\frac{-b+\omega }{2a}=\frac{-\left( 1+i \right)+\sqrt{3}\left( 1-i \right)}{2}=\frac{\sqrt{3}-1}{2}-i\frac{1+\sqrt{3}}{2}$

${{z}_{2}}=\frac{-b-\omega }{2a}=\frac{-\left( 1+i \right)+\sqrt{3}\left( -1+i \right)}{2}=\frac{-\sqrt{3}-1}{2}+i\frac{-1+\sqrt{3}}{2}$