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Question Number 76306 by Hassen_Timol last updated on 26/Dec/19

$$z=a+bi$$$$Z=\frac{-7+z}{-3+iz}$$$$$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{points}\mathrm{M}$$$$\mathrm{of}\mathrm{coordonates}(a,b)\mathrm{such}\mathrm{as}Z\mathrm{is}\mathrm{real}?$$

Answered by mr W last updated on 26/Dec/19

$$Z=\frac{-7+a+bi}{-3+i(a+bi)}$$$$=\frac{(-7+a)+bi}{(-3-b)+ai}$$$$=\frac{[(-7+a)+bi][(-3-b)-ai]}{[(-3-b)+ai][(-3-b)-ai]}$$$$=\frac{(-7+a)(-3-b)+ab+[(-3-b)b-(-7+a)a]i}{{(-3-b)}^2+a^2}$$$$=\frac{21-3a+7b+(7a-3b-a^2-b^2)i}{{(-3-b)}^2+a^2}$$$$suchthatZisreal,$$$$7a-3b-a^2-b^2=0$$$$\Rightarrow {(a-\frac{7}{2})}^2+{(b+\frac{3}{2})}^2={\left(\frac{7}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2$$$$\Rightarrow {(a-\frac{7}{2})}^2+{(b+\frac{3}{2})}^2={\left(\frac{\sqrt{58}}{2}\right)}^2$$$$locusofMisacirclewithradius$$$$\frac{\sqrt{58}}{2}andcenterat(\frac{7}{2},-\frac{3}{2}).$$

Commented by Hassen_Timol last updated on 26/Dec/19

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir},\mathrm{may}\mathrm{God}\mathrm{bless}\mathrm{you}!$$

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