Question Number 76306 by Hassen_Timol last updated on 26/Dec/19
$${z}\:=\:{a}\:+\:{bi} \\ $$$${Z}\:=\:\frac{−\mathrm{7}\:+\:{z}}{−\mathrm{3}\:+\:{iz}} \\ $$$$ \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{points}\:\mathrm{M} \\ $$$$\mathrm{of}\:\mathrm{coordonates}\:\left({a},{b}\right)\:\mathrm{such}\:\mathrm{as}\:{Z}\:\mathrm{is}\:\mathrm{real}\:? \\ $$
Answered by mr W last updated on 26/Dec/19
![Z=((−7+a+bi)/(−3+i(a+bi))) =(((−7+a)+bi)/((−3−b)+ai)) =(([(−7+a)+bi][(−3−b)−ai])/([(−3−b)+ai][(−3−b)−ai])) =(((−7+a)(−3−b)+ab+[(−3−b)b−(−7+a)a]i)/((−3−b)^2 +a^2 )) =((21−3a+7b+(7a−3b−a^2 −b^2 )i)/((−3−b)^2 +a^2 )) such that Z is real, 7a−3b−a^2 −b^2 =0 ⇒(a−(7/2))^2 +(b+(3/2))^2 =((7/2))^2 +((3/2))^2 ⇒(a−(7/2))^2 +(b+(3/2))^2 =(((√(58))/2))^2 locus of M is a circle with radius ((√(58))/2) and center at ((7/2),−(3/2)).](https://www.tinkutara.com/question/Q76308.png)
$${Z}=\frac{−\mathrm{7}+{a}+{bi}}{−\mathrm{3}+{i}\left({a}+{bi}\right)} \\ $$$$=\frac{\left(−\mathrm{7}+{a}\right)+{bi}}{\left(−\mathrm{3}−{b}\right)+{ai}} \\ $$$$=\frac{\left[\left(−\mathrm{7}+{a}\right)+{bi}\right]\left[\left(−\mathrm{3}−{b}\right)−{ai}\right]}{\left[\left(−\mathrm{3}−{b}\right)+{ai}\right]\left[\left(−\mathrm{3}−{b}\right)−{ai}\right]} \\ $$$$=\frac{\left(−\mathrm{7}+{a}\right)\left(−\mathrm{3}−{b}\right)+{ab}+\left[\left(−\mathrm{3}−{b}\right){b}−\left(−\mathrm{7}+{a}\right){a}\right]{i}}{\left(−\mathrm{3}−{b}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{21}−\mathrm{3}{a}+\mathrm{7}{b}+\left(\mathrm{7}{a}−\mathrm{3}{b}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){i}}{\left(−\mathrm{3}−{b}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$${such}\:{that}\:{Z}\:{is}\:{real}, \\ $$$$\mathrm{7}{a}−\mathrm{3}{b}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({a}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({b}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({a}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({b}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{58}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${locus}\:{of}\:{M}\:{is}\:{a}\:{circle}\:{with}\:{radius} \\ $$$$\frac{\sqrt{\mathrm{58}}}{\mathrm{2}}\:{and}\:{center}\:{at}\:\left(\frac{\mathrm{7}}{\mathrm{2}},−\frac{\mathrm{3}}{\mathrm{2}}\right). \\ $$
Commented by Hassen_Timol last updated on 26/Dec/19
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir},\:\mathrm{may}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}! \\ $$