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Question-175154




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

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