Complex Exercise Asked by KL liu in the 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$\left( {{y}^{4}}-{{5}^{2}} \right)\left( {{y}^{4}}+{{5}^{2}} \right)=0$

$\Leftrightarrow \left( {{y}^{2}}-5 \right)\left( {{y}^{2}}+5 \right)\left( {{y}^{4}}+{{5}^{2}} \right)=0$

So ${{y}^{2}}=5\,\,or\,\,{{y}^{2}}=-5\,\,or\,\,{{y}^{4}}=-{{5}^{2}}$

Hence ${{y}^{2}}=5\Leftrightarrow y=\pm \sqrt{5}\,$ , ${{y}^{2}}=-5\Leftrightarrow y=\pm i\sqrt{5}$ and

${{y}^{4}}=-{{5}^{2}}$ Take $y=r{{e}^{i\theta }}\Leftrightarrow {{y}^{4}}={{r}^{4}}{{e}^{i4\theta }}={{5}^{2}}{{e}^{i\pi }}$

So ${{r}^{4}}={{5}^{2}}\,\And \,4\theta =\pi +2k\pi $ $\Rightarrow r=\sqrt{5}\,\,\And \,\,\theta =\frac{\pi }{4}+\frac{k\pi }{2}$ hence \(y=\sqrt{5}{{e}^{i\frac{\pi }{4}+i\frac{k\pi }{2}}}\,,\,\,\,\,0\le k\le 3\)

${{y}^{4}}=-{{5}^{2}}\Leftrightarrow {{y}^{4}}={{i}^{2}}{{5}^{2}}\Leftrightarrow y=\pm \sqrt[4]{{{i}^{2}}{{5}^{2}}}=\pm {{i}^{2/4}}{{5}^{2/4}}=\pm {{\left( -1 \right)}^{1/4}}\sqrt{5}$

Now by backward substitution we get :

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

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

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

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

$y={{\left( -1 \right)}^{1/4}}\sqrt{5}\Leftrightarrow x=\sqrt[3]{5{{\left( -1 \right)}^{1/4}}\sqrt{5}}={{\left( -1 \right)}^{3/4}}{{\left( {{\left( {{5}^{3}} \right)}^{1/2}} \right)}^{1/3}}={{\left( -1 \right)}^{3/4}}\sqrt{5}$

$y=-{{\left( -1 \right)}^{1/4}}\sqrt{5}\Rightarrow x=\sqrt[3]{-5{{\left( -1 \right)}^{1/4}}\sqrt{5}}=\sqrt[3]{{{i}^{2}}{{i}^{2/4}}\sqrt{{{5}^{3}}}}=\sqrt[3]{{{i}^{5/2}}{{5}^{3/2}}}={{i}^{5/6}}\sqrt{5}=-{{\left( -1 \right)}^{3/4}}\sqrt{5}$

$y={{\left( -1 \right)}^{3/4}}\sqrt{5}\Leftrightarrow x={{\left( -1 \right)}^{1/4}}\sqrt{5}$ and $y=-{{\left( -1 \right)}^{3/4}}\sqrt{5}\Leftrightarrow x=-{{\left( -1 \right)}^{1/4}}\sqrt{5}$

So the solution set is :

$S=\left\{ \begin{align}
  & \left( 0,0 \right),\left( \sqrt{5},\sqrt{5} \right),\left( -\sqrt{5},-\sqrt{5} \right),\left( i\sqrt{5},-i\sqrt{5} \right),\left( -i\sqrt{5},i\sqrt{5} \right),\left( {{\left( -1 \right)}^{1/4}}\sqrt{5},{{\left( -1 \right)}^{3/4}}\sqrt{5} \right), \\
 & \left( -{{\left( -1 \right)}^{1/4}}\sqrt{5},-{{\left( -1 \right)}^{3/4}}\sqrt{5} \right),\left( {{\left( -1 \right)}^{3/4}}\sqrt{5},{{\left( -1 \right)}^{1/4}}\sqrt{5} \right),\left( -{{\left( -1 \right)}^{3/4}}\sqrt{5},-{{\left( -1 \right)}^{1/4}}\sqrt{5} \right) \\
\end{align} \right\}$