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Question Number 130549 by EDWIN88 last updated on 26/Jan/21

$$thesolutionofequation$$$$\mid z\mid -z=1+2iis\mathrm{\_}\mathrm{\_}$$

Answered by Dwaipayan Shikari last updated on 26/Jan/21

$$z=x+iy$$$$\sqrt{x^2+y^2}=(x+1)+i(y+2)$$$$y=-2$$$$x+1=\sqrt{x^2+4}\Rightarrow x=\frac{3}{2}$$$$z=\frac{3}{2}-2i$$

Answered by liberty last updated on 26/Jan/21

$$\mid \mathrm{z}\mid =\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\Rightarrow \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}-\mathrm{x}-\mathrm{y}i=1+2i$$$$\{\begin{array}{l}y=-2\\ \sqrt{x^2+4}-x=1;\sqrt{{\mathrm{x}}^2+4}=1+\mathrm{x}\end{array}$$$$\Rightarrow {\mathrm{x}}^2+4=1+2\mathrm{x}+{\mathrm{x}}^2;\mathrm{x}=\frac{3}{2}$$$$\therefore \mathrm{z}=\frac{3}{2}-2i$$

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