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Question Number 139116 by EnterUsername last updated on 22/Apr/21
The number of solutions to the equation  z(z−2i)^(−) =2(z+i) is ?
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{z}\overline {\left(\mathrm{z}−\mathrm{2i}\right)}=\mathrm{2}\left(\mathrm{z}+\mathrm{i}\right)\:\mathrm{is}\:? \\ $$
Answered by qaz last updated on 22/Apr/21
z∙z−2i^(−) =zz^− +z−2i^(−) =zz^− +2iz=2(z+i)  z=a+bi  a^2 +b^2 +2i(a+bi)=2(a+(b+1)i)  ⇒a^2 +b^2 −2b=2a       2a=2(b+1)  ⇒b=((1+(√3))/2),a=((3+(√3))/2)  or  b=((1−(√3))/2),a=((3−(√3))/2)  ⇒z=((3+(√3))/2)+((1+(√3))/2)i   or     ((3−(√3))/2)+((1−(√3))/2)i
$${z}\centerdot\overline {{z}−\mathrm{2}{i}}={z}\overset{−} {{z}}+{z}\overline {−\mathrm{2}{i}}={z}\overset{−} {{z}}+\mathrm{2}{iz}=\mathrm{2}\left({z}+{i}\right) \\ $$$${z}={a}+{bi} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{i}\left({a}+{bi}\right)=\mathrm{2}\left({a}+\left({b}+\mathrm{1}\right){i}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{b}=\mathrm{2}{a} \\ $$$$\:\:\:\:\:\mathrm{2}{a}=\mathrm{2}\left({b}+\mathrm{1}\right) \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}},{a}=\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${or}\:\:{b}=\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}},{a}=\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow{z}=\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\:\:\:{or}\:\:\:\:\:\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$

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