Complex Exercise asked by kl liu in Department of mathematics group

Exercise:

Solve in $\mathbb{C},$ ${{x}^{3}}=5y\,\,\And \,\,{{y}^{3}}=5x$

Solution: we have ${{y}^{3}}=5x\,\,\And \,\,{{x}^{3}}=5y\Rightarrow x=\sqrt[3]{5y}$

So ${{y}^{3}}=5\sqrt[3]{5y}$ $\Rightarrow {{y}^{9}}={{5}^{3}}\times 5y\Rightarrow {{y}^{9}}-{{5}^{4}}y=0\Rightarrow y\left( {{y}^{8}}-{{5}^{4}} \right)=0$

Hence $y=0\,\,\,\,or\,\,\,{{y}^{8}}-{{5}^{4}}=0$ thus ${{y}^{8}}={{5}^{4}}\Rightarrow \sqrt[4]{{{y}^{8}}}=\pm \sqrt[4]{{{5}^{4}}}\Rightarrow {{y}^{2}}=\pm 5$

So we have two cases to be considered

Case 1: ${{y}^{2}}=5\Leftrightarrow y=\pm \sqrt{5}$

Case 2: ${{y}^{2}}=-5\Leftrightarrow y=\pm \sqrt{-5}=\pm \sqrt{5{{i}^{2}}}=\pm i\sqrt{5}$

Now by backward substitution we get :

$y=\sqrt{5}\Rightarrow x=\sqrt[3]{5\sqrt{5}}=\sqrt[3]{\sqrt{{{5}^{3}}}}=\sqrt[3]{{{5}^{3/2}}}={{\left( {{5}^{3/2}} \right)}^{1/3}}={{5}^{1/2}}=\sqrt{5}$

$y=-\sqrt{5}\Rightarrow x=\sqrt[3]{-5\sqrt{5}}={{\left( -1 \right)}^{1/3}}\sqrt{5}$

$y=i\sqrt{5}\Rightarrow x=\sqrt[3]{5i\sqrt{5}}=\sqrt[3]{\sqrt{{{5}^{3}}{{i}^{2}}}}=\sqrt[3]{{{\left( {{5}^{3}}{{i}^{2}} \right)}^{1/2}}}={{\left( {{\left( {{5}^{3}}{{i}^{2}} \right)}^{1/2}} \right)}^{1/3}}={{\left( -1 \right)}^{1/6}}\sqrt{5}$

$y=-i\sqrt{5}\Rightarrow x=\sqrt[3]{-5i\sqrt{5}}=\sqrt[3]{{{i}^{3}}\sqrt{{{5}^{3}}}}=-{{\left( -1 \right)}^{5/6}}\sqrt{5}$

So the solution set is :

$S=\left\{ \left( 0,0 \right),\left( \sqrt{5},\sqrt{5} \right),\left( -\sqrt{5},-\sqrt{5} \right),\left( i\sqrt{5},-i\sqrt{5} \right),\left( -i\sqrt{5},i\sqrt{5} \right) \right\}$