Question Number 177702 by lapache last updated on 08/Oct/22
$${Solve}\:{in}\:\mathbb{C} \\ $$$$\begin{cases}{{x}+{y}={z}}\\{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{{z}}}\end{cases} \\ $$
Answered by Frix last updated on 08/Oct/22
$$\left(\mathrm{1}\right)\:{x}={z}−{y} \\ $$$$\left(\mathrm{2}\right)\:{x}=\frac{{yz}}{{y}−{z}} \\ $$$$\Leftrightarrow \\ $$$${z}−{y}=\frac{{yz}}{{y}−{z}} \\ $$$${y}^{\mathrm{2}} +{yz}−{z}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{y}={z}×\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$$$\Rightarrow \\ $$$${x}={z}×\left(\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$$${z}\in\mathbb{C}\:\Rightarrow\:{z}={p}+{q}\mathrm{i} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:=\begin{pmatrix}{\left({p}+{q}\mathrm{i}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)}\\{\left({p}+{q}\mathrm{i}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)}\\{{p}+{q}\mathrm{i}}\end{pmatrix} \\ $$
Commented by Tawa11 last updated on 08/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 08/Oct/22
$$\frac{{x}+{y}}{{xy}}=\frac{\mathrm{1}}{{z}} \\ $$$$\Rightarrow{xy}=\left({x}+{y}\right){z}={z}^{\mathrm{2}} \\ $$$${x}+{y}={z} \\ $$$$\Rightarrow{x},{y}\:{are}\:{roots}\:{of}\:{t}^{\mathrm{2}} −{zt}+{z}^{\mathrm{2}} =\mathrm{0} \\ $$$${x},\:{y}=\frac{{z}\pm\sqrt{{z}^{\mathrm{2}} −\mathrm{4}{z}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\left(\mathrm{1}\pm\mathrm{3}{i}\right){z}}{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 08/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$