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Problem: Describe the locus $\mathcal{L}$ determined by the equation $$z(\overline{z}+2)=3, z \in \mathbb{C}.$$

My argument: Let $z=x+iy$. We compute as follows:

$$(x+iy)(x-iy+2)=3;$$ $$(x^2+2x+y^2)+(2y)i=3;$$

equaling the real and imaginary parts, we get the system of equations \begin{cases} x^2+2x+y^2=3\\ 2y=0 \end{cases}

Solving it, I conclude that $\mathcal{L}=\{(-3,0),(1,0)\}$.

Person B's argument: By completing squares in the equation $$(x^2+2x+y^2)+(2y)i=3$$ they get $$(x+1)^2+(y+i)^2=3.$$ Person B's conclusion: the locus is the circumference with center $(-1, -i)$ and radius 3^(1/2).

Now, I don't even understand what their conclusion means. How can a circumference in $\mathbb{R}^2$ (which is isomorphic to $\mathbb{C}$) have the point $(-1, -i) \in \mathbb{R} \times \mathbb{C}$ as its center? Person B argued that "since we are working with complex numbers we can work with complex coordinates", but for me this as wrong as talking about a complex number with real or imaginary part being a complex number as well (instead of a real number).

My question: Is Person B's argument wrong? How will you make Person B realize their mistake?

Amelian
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  • @DonThousand a way of constructing $\mathbb{C} $ is by defining a sum and a product on the set $\mathbb{R} \times \mathbb{R} $; a complex number has a unique representation as an ordered pair $(a,b)$ with $a$ and $b$ real numbers. – Amelian Aug 09 '21 at 16:13
  • I'm aware. I've seen $(a,b)$ written as $(a,bi)$ sometimes. Can't say I like the notation, but it does exist. – Rushabh Mehta Aug 09 '21 at 16:14
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    The essential failure in the second argument is that Person B has expanded the domain of $x$ and $y$. When we say 'let $z=x+iy$' there's an implicit assumption that $x,y\in\mathbb{R}$; otherwise we can't assign each $z$ to a unique $x,y$ pair. So while $(x+1)^2+(y+i)^2=3$ is a correct way of writing the equation that you get, we have to restrict $x$ and $y$ to be in $\mathbb{R}$ — and that takes us back to the first solution. – Steven Stadnicki Aug 09 '21 at 16:14
  • What happened to the $-ixy$ term? – user58697 Aug 09 '21 at 17:12
  • @steven -- Feel the rhythm, feel the rhyme, get on up! It's answer time! Would like to see your comment so converted. – Oscar Lanzi Aug 09 '21 at 17:56

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As $z(\bar z+2)$ is real it equals its conjugate $\bar z(z+2)$ from which $\bar z=z$ follows. Solving $z(z+2)=3$ gives $z=-3$ or $z=1$.

Michael Hoppe
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