Question Number 175154 by Shrinava last updated on 21/Aug/22
Commented by mr W last updated on 21/Aug/22
$${i}\:{don}'{t}\:{think}\:{there}\:{are}\:{unique}\:{solutions}. \\ $$
Answered by MJS_new last updated on 21/Aug/22
$$\mathrm{we}\:\mathrm{are}\:\mathrm{free}\:\mathrm{to}\:\mathrm{choose}\:\mathrm{2}\:\mathrm{out}\:\mathrm{if}\:{x},\:{y},\:{z}\:\mathrm{and}\:\mathrm{solve} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{one} \\ $$$${x}=\frac{\mathrm{2}{y}\left({y}+{z}\right){z}}{\left({y}^{\mathrm{2}} +\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)}\pm\frac{{y}^{\mathrm{2}} {z}^{\mathrm{2}} −\left({y}+{z}\right)^{\mathrm{2}} −\mathrm{1}}{\left({y}^{\mathrm{2}} +\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{i} \\ $$$$\mathrm{with}\:{y},\:{z}\:\in\mathbb{C}\backslash\left\{\pm\mathrm{i}\right\} \\ $$