# Describe the loci: $a|z|^2 + kz + \overline {kz} + d = 0$

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!

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

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 $$|k|^2>ad$$.

• 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. Jan 14, 2023 at 23:38
• Hi John Omielan, thank you for pointing out this to me! I didn't think about this possibility. Jan 15, 2023 at 1:31
• 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? Jan 15, 2023 at 1:35
• If $z=x+yi$ and $k=a+bi$, $2\text{Re}(kz)+d=0\implies 2ax-2by+d=0$. Jan 15, 2023 at 1:39