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Question Number 176513 by CrispyXYZ last updated on 20/Sep/22

$$z\in \mathbb{C},\frac{z-3\mathrm{i}}{z+\mathrm{i}}\in {\mathbb{R}}^{-},\frac{z-3}{z+1}\in \mathbb{I}$$$$\mathrm{find}z.$$

Answered by a.lgnaoui last updated on 20/Sep/22

$$\mathrm{Posons}\mathrm{z}=\mathrm{x}+\mathrm{iy}$$$$\frac{\mathrm{z}-3\mathrm{i}}{\mathrm{z}+\mathrm{i}}=\frac{\mathrm{x}+(\mathrm{y}-3)\mathrm{i}}{\mathrm{x}+(\mathrm{y}+1)\mathrm{i}}=\frac{[\mathrm{x}+(\mathrm{y}-3)\mathrm{i}][\mathrm{x}-(\mathrm{y}+1)\mathrm{i}]}{{\mathrm{x}}^2+{(\mathrm{y}+1)}^2}=\frac{[{\mathrm{x}}^2+(\mathrm{y}+1)(\mathrm{y}-3)]+\left[(\mathrm{x}(\mathrm{y}-3)-\mathrm{x}(\mathrm{y}+1))\mathrm{i}\right]}{{\mathrm{x}}^2-{(\mathrm{y}+1)}^2}$$$$$$$$\frac{\mathrm{z}-3\mathrm{i}}{\mathrm{z}+\mathrm{i}}=\frac{({\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{y}-3)-\left(4\mathrm{x}\right)\mathrm{i}}{{\mathrm{x}}^2+{(\mathrm{y}+1)}^2}$$$$$$$$\frac{\mathrm{z}-3}{\mathrm{z}+1}=\frac{(\mathrm{x}-3)+\mathrm{yi}}{(\mathrm{x}+1)+\mathrm{yi}}=\frac{[(\mathrm{x}-3)+\mathrm{yi}][(\mathrm{x}+1)-\mathrm{yi}]}{{(\mathrm{x}+1)}^2+{\mathrm{y}}^2}=\frac{[(\mathrm{x}-3)(\mathrm{x}+1)+{\mathrm{y}}^2]+[((\mathrm{x}+1)\mathrm{y}-(\mathrm{x}-3)\mathrm{y})\mathrm{i}}{{(\mathrm{x}+1)}^2+{\mathrm{y}}^2}$$$$$$$$\frac{\mathrm{z}-3}{\mathrm{z}+1}=\frac{[{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-3]+\left(4y\right)i}{{(\mathrm{x}+1)}^2+{\mathrm{y}}^2}$$$$$$$$\frac{\mathrm{z}-3\mathrm{i}}{\mathrm{z}+\mathrm{i}}\in {\mathbb{R}}^{-}\Rightarrow \mathrm{x}=0\frac{\mathrm{z}-3\mathrm{i}}{\mathrm{z}+\mathrm{i}}=\frac{{\mathrm{y}}^2-2\mathrm{y}-3}{{(\mathrm{y}+1)}^2}<0$$$$\frac{\mathrm{z}-3}{\mathrm{z}+1}\mathrm{imaginaire}{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-3=0(\mathrm{y}\ne 0)\mathrm{avec}\mathrm{x}=0$$$$donc\mathrm{z}\mathrm{verifie}$$$$\mathrm{y}=\pm \sqrt{3}\mathrm{y}=-\sqrt{3}\mathrm{rejete}$$$$\mathrm{y}=+\sqrt{3}$$$$\frac{\mathrm{z}-3\mathrm{i}}{\mathrm{z}+\mathrm{i}}=\frac{3-2\sqrt{3}-3}{{(\sqrt{3}+1)}^2}<0\mathrm{et}$$$$\frac{\mathrm{z}-3}{\mathrm{z}+1}=\frac{4\sqrt{3}}{3}\mathrm{i}\in \mathbf{I}\Rightarrow \mathbf{z}=\sqrt{3}\mathbf{i}$$$$$$

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