complex | aussiemathstutor

$\displaystyle \begin{array}{l}{{z}^{4}}-2{{z}^{2}}+4=0,z\in C\\\left( {{{z}^{4}}-2{{z}^{2}}-2{{z}^{2}}+4} \right)+2{{z}^{2}}=0\\\left( {{{z}^{2}}-2} \right)+2{{z}^{2}}=0\\{{\left( {{{z}^{2}}-2} \right)}^{2}}-{{\left( {i\sqrt{2}z} \right)}^{2}}=0\\\left( {{{z}^{2}}-i\sqrt{2}z-2} \right)\left( {{{z}^{2}}+i\sqrt{2}z-2} \right)=0\\{{z}^{2}}-i\sqrt{2}z-2=0\\a=1,b=-i\sqrt{2},c=-2\\z=\frac{{-\left( {-i\sqrt{2}} \right)\pm \sqrt{{{{{\left( {-i\sqrt{2}} \right)}}^{2}}-4\left( 1 \right)\left( {-2} \right)}}}}{{2\left( 1 \right)}}\\=\frac{{i\sqrt{2}\pm \sqrt{6}}}{2}\\=\frac{{\pm \sqrt{6}}}{2}+\frac{{i\sqrt{2}}}{2}\\{{z}^{2}}+i\sqrt{2}z-2=0\\a=1,b=i\sqrt{2},c=-2\\z=\frac{{-\left( {i\sqrt{2}} \right)\pm \sqrt{{{{{\left( {i\sqrt{2}} \right)}}^{2}}-4\left( 1 \right)\left( {-2} \right)}}}}{{2\left( 1 \right)}}\\=\frac{{-i\sqrt{2}\pm \sqrt{6}}}{2}\\=\frac{{\pm \sqrt{6}}}{2}-\frac{{i\sqrt{2}}}{2}\\z=\left( {\frac{{\pm \sqrt{6}}}{2}+\frac{{i\sqrt{2}}}{2}} \right),\left( {\frac{{\pm \sqrt{6}}}{2}-\frac{{i\sqrt{2}}}{2}} \right)\end{array}$

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Complex solve using complete the square variation.