I need to find the loci that $a|z|^2 + kz + \overline {kz} + d = 0$ represents. I'm also given the condition that $k \in \mathbb{C}$ and $a,d \in \mathbb{R}$ and $|k^2| > ad$. Then I first divide both sides of the equation by $a$. Then I get $|z|^2 + \frac{kz}{a} + \frac{\overline{kz}}{a} + \frac{d}{a} = 0$. Then I used one property of complex conjugate to simplify this and I get$$|z|^2 + \frac{2}{a} Re(kz) + \frac{d}{a} = 0$$ Then I don't know how to further analyze this equation, and I feel like I haven't used the condition that $|k^2| > ad$ and I'm confused about where to use it. Thanks!

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    $\begingroup$ Don't you think it is a circle? I get $4ad<|k|^2$ as the condition. $\endgroup$
    – GEdgar
    Jan 14, 2023 at 22:27
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    $\begingroup$ Which one is correct, $\overline{k}z$ in the title or $\overline{kz}$ in the body? $\endgroup$
    – user376343
    Jan 14, 2023 at 22:33
  • $\begingroup$ I'm sorry that I got the title wrong. $\overline {kz}$ in the body is correct. $\endgroup$
    – Wendy Lyu
    Jan 14, 2023 at 22:40
  • $\begingroup$ Yeah, I also guess thats It's a circle. But I'm just stuck on how to give a mathematical expression of this circle. $\endgroup$
    – Wendy Lyu
    Jan 14, 2023 at 22:41
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    $\begingroup$ Does this answer your question? The equation of a circle on a complex plane? $\endgroup$
    – Jean Marie
    Jan 15, 2023 at 1:26

1 Answer 1


Note that

$$a\left|z+\frac{\bar k}{a}\right|^2= a|z|^2+ kz +\overline{kz}+\frac{|k|^2}{a},$$

so the equation can be rewritten as

$$a\left|z+\frac{\bar k}{a}\right|^2 = -d +\frac{|k|^2}{a}.$$

This is a circle centered at $-\bar k/a$ with positive radius as $|k|^2>ad$.

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    $\begingroup$ Note that as I commented to the other answer, the question doesn't explicitly state that $a \neq 0$, with both the OP and your answer not dealing with the possibility that $a = 0$. Note that the result would then be a straight line, but that can be considered to be a degenerate form of a circle. $\endgroup$ Jan 14, 2023 at 23:38
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    $\begingroup$ Hi John Omielan, thank you for pointing out this to me! I didn't think about this possibility. $\endgroup$
    – Wendy Lyu
    Jan 15, 2023 at 1:31
  • $\begingroup$ Hi John, I did'nt quite get it why it would be a line when a = 0. Could you give me a brief explanation please? $\endgroup$
    – Wendy Lyu
    Jan 15, 2023 at 1:35
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    $\begingroup$ If $z=x+yi$ and $k=a+bi$, $2\text{Re}(kz)+d=0\implies 2ax-2by+d=0$. $\endgroup$
    – user51547
    Jan 15, 2023 at 1:39

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