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Show that the locus of points $z$ in the complex plane satisfying an equation of the type $Az\bar{z}+Bz+\bar{B}\bar{z}+c=0$, in which both $A$ and $C$ are real numbers and $|B|^2-AC>0$, is a circle when $A\neq0$ and a line otherwise. Conversely, show that any circle or line in the complex plane admits an equation of the kind just described.

$Az\bar{z}+Bz+\bar{B}\bar{z}+C=0 \Leftrightarrow A\left | z \right |^2+2Re(Bz)+C=0\Leftrightarrow \left | z \right |^2+2Re(Bz)/A+C/A=0\Leftrightarrow \left | z \right |^2=-2Re(Bz)-C/A$, but here $-2Re(Bz)-C/A\geq 0$? Why $|B|^2-AC>0$? If $A=0\Rightarrow 2Re(Bz)+C=0\Rightarrow Re(Bz)=-C/2$ How do I conclude that this is a straight line? How would you make the "conversely"? Thank you very much for your help.

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user425181
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2 Answers2

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Hints.

  • If $A \ne 0\,$:  multiply the equation by $A$, then: $$ A^2z\bar{z}+ABz+A\bar{B}\bar{z}\color{red}{+ B \bar B - B \bar B}+AC = 0 \\ \iff\;\; (Az+\bar B)(A\bar z + B) = B \bar B -AC \\ \iff\;\; \left|Az+\bar B\right|^2 = |B|^2 - AC $$ The latter equation says that the distance from $z$ to the fixed point $-\bar B / A$ is constant.

  • If $A=0\,$: $\;\operatorname{Re}(Bz)=−C/2\,$ iff $\,Bz\,$ lies on a vertical line through point $-C/2\,$ on the real axis.

dxiv
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If $z=x+iy$ and $B=B_1+iB_2$, then $$Az\bar{z}+Bz+\bar{B}\bar{z}+C=0\\\Longleftrightarrow Ax^2+Ay^2+2B_1x-2B_2y+C=0$$ This is a line if $A=0$. If $A\neq 0$, we can rearrange it as $$\left(x+\dfrac{B_1}{A}\right)^2+\left(y-\dfrac{B_2}{A}\right)^2=\dfrac{B_1^2+B_2^2-CA}{A}=\dfrac{|B|^2-CA}{A}$$ which is a circle when $|B|^2-CA\neq 0$.

Prajwal Kansakar
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