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Question Number 175154 by Shrinava last updated on 21/Aug/22

Commented by mr W last updated on 21/Aug/22

$$i{don}^{\prime }tthinkthereareuniquesolutions.$$

Answered by MJS_new last updated on 21/Aug/22

$$\mathrm{we}\mathrm{are}\mathrm{free}\mathrm{to}\mathrm{choose}2\mathrm{out}\mathrm{if}x,y,z\mathrm{and}\mathrm{solve}$$$$\mathrm{for}\mathrm{the}\mathrm{remaining}\mathrm{one}$$$$x=\frac{2y(y+z)z}{(y^2+1)(z^2+1)}\pm \frac{y^2{z}^2-{(y+z)}^2-1}{(y^2+1)(z^2+1)}\mathrm{i}$$$$\mathrm{with}y,z\in \mathbb{C}\mathrm{\setminus }\{\pm \mathrm{i}\}$$

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