February | 2015 | aussiemathstutor

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

$\begin{array}{l}{{z}^{4}}-2{{z}^{2}}+4=0,z\in C\\\text{Let }w={{z}^{2}}\Rightarrow \\{{w}^{2}}-2w+4=0\\\left( {{{w}^{2}}-2w+1} \right)+4-1=0\\{{\left( {w-1} \right)}^{2}}-3=0\\{{\left( {w-1} \right)}^{2}}-\left( {i\sqrt{3}} \right)=0\\\left( {w-1-i\sqrt{3}} \right)\left( {w-1+i\sqrt{3}} \right)=0\\w-1-i\sqrt{3}=0\Rightarrow \\{{z}^{2}}=1+i\sqrt{3}=2cis\left( {\frac{{\pi \left( {6n+1} \right)}}{3}} \right)\\z=\sqrt{2}cis\left( {\frac{{\pi \left( {6n+1} \right)}}{6}} \right)\\n=0,{{z}_{1}}=\sqrt{2}cis\left( {\frac{{\pi \left( 1 \right)}}{6}} \right)=\sqrt{2}\left( {\frac{{\sqrt{3}}}{2}+i\frac{1}{2}} \right)\\n=1,{{z}_{2}}=\sqrt{2}cis\left( {\frac{{\pi \left( 7 \right)}}{6}} \right)=\sqrt{2}\left( {-\frac{{\sqrt{3}}}{2}-i\frac{1}{2}} \right)\\w-1+i\sqrt{3}=0\Rightarrow \\{{z}^{2}}=1-i\sqrt{3}=2cis\left( {\frac{{\pi \left( {6n-1} \right)}}{3}} \right)\\z=\sqrt{2}cis\left( {\frac{{\pi \left( {6n-1} \right)}}{6}} \right)\\n=0,{{z}_{1}}=\sqrt{2}cis\left( {\frac{{\pi \left( {-1} \right)}}{6}} \right)=\sqrt{2}\left( {\frac{{\sqrt{3}}}{2}-i\frac{1}{2}} \right)\\n=1,{{z}_{2}}=\sqrt{2}cis\left( {\frac{{\pi \left( 5 \right)}}{6}} \right)=\sqrt{2}\left( {-\frac{{\sqrt{3}}}{2}+i\frac{1}{2}} \right)\end{array}$

aussiemathstutor

Home of the Aussie Maths Tutor.

Monthly Archives: February 2015

Complex solve using complete the square variation.

Complex Solve 1