Question Number 11334 by uni last updated on 20/Mar/17
![z+∣z∣=9−3i⇒Re(z)](https://www.tinkutara.com/question/Q11334.png)
$$\mathrm{z}+\mid\mathrm{z}\mid=\mathrm{9}−\mathrm{3i}\Rightarrow\mathrm{Re}\left(\mathrm{z}\right) \\ $$
Answered by mrW1 last updated on 20/Mar/17
![z=a+bi ∣z∣=(√(a^2 +b^2 )) z+∣z∣=(a+(√(a^2 +b^2 )))+bi=9−3i ⇒b=−3 a+(√(a^2 +b^2 ))=9 a+(√(a^2 +9))=9 a^2 +9=(9−a)^2 a^2 +9=81+a^2 −18a a=((72)/(18))=4 ⇒Re(z)=a=4](https://www.tinkutara.com/question/Q11335.png)
$${z}={a}+{bi} \\ $$$$\mid{z}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${z}+\mid{z}\mid=\left({a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)+{bi}=\mathrm{9}−\mathrm{3}{i} \\ $$$$\Rightarrow{b}=−\mathrm{3} \\ $$$${a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{9} \\ $$$${a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{9}}=\mathrm{9} \\ $$$${a}^{\mathrm{2}} +\mathrm{9}=\left(\mathrm{9}−{a}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +\mathrm{9}=\mathrm{81}+{a}^{\mathrm{2}} −\mathrm{18}{a} \\ $$$${a}=\frac{\mathrm{72}}{\mathrm{18}}=\mathrm{4} \\ $$$$\Rightarrow{Re}\left({z}\right)={a}=\mathrm{4} \\ $$$$ \\ $$