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When describing sets of points in the complex plane such as $|2z-i| =4$ would it be correct to describe the set exactly as we would in the $x,y$ plane?

For instance, Letting $z=x+iy$ for some $x,y \in \mathbb{R}$

yields the equation $x^2+(y-\frac{1}{2})^2=4$

And would it be correct to say this is the circle in the complex plane described by $x^2+(y-\frac{1}{2})^2=4$

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  • $\begingroup$ Do you know Euler's formula? It provides for an identification of $\Bbb R^2$ and $\Bbb C$. It's $e^{i\theta}=\cos\theta+i\sin\theta$. Then every $(x,y)$ can be identified with $(r,\theta)$ via $z=re^{i\theta}$.' The coordinates $(r,\theta)$ are called "polar coordinates". $\endgroup$ – Chris Custer Jan 22 at 3:36
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Would it be correct to describe the set exactly as we would in the x,y plane?

If there is no more structure given to $\mathbb{R}$ and $\mathbb{C}$ else than the norms, I will say yes, since we have the isomorphism $iso:\mathbb{C}\ni z=x+iy\mapsto (x,y)\in\mathbb{R}^2$ (bijection that conserving norm).

And would it be correct to say this ($|2z−i|=4$) is the circle in the complex plane described by $x^2+(y−\frac{1}{2})^2=4$

Yes, in sense that $$iso(\{z\in\mathbb{Z}:|2z-i|=4\})=\{(x,y)\in\mathbb{R}^2:x^2+(y-\frac{1}{2})^2=4\}.$$

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  • $\begingroup$ Is the $i$ in front of the first set a mistake? $\endgroup$ – 68e1515 Jan 22 at 2:48
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    $\begingroup$ @68e1515 Sorry, it may be misleading notation. I fixed it. It means the image of the given set by isomorphism from $\mathbb{C}$ to $\mathbb{R}^2$ can be uniquely specified as a set of $\mathbb{R}^2$. $\endgroup$ – An Jin Jan 22 at 2:54
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I would say, rather, that this is the circle in the complex plane analogous to $\displaystyle x^2+\left(y-\frac{1}{2}\right)^2=4$.

The reason is, $\left |z-\frac{i}{2}\right|=2$ is well understood to mean a circle with radius $1$ and center $z=\frac{1}{2}$, and that the complex plane, as you note, uses $z$ instead of $x$ and $y$.

(Distance from center = radius)

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  • $\begingroup$ What if the equation in the complex plane is $|z-1|=|z+i|$, can we say this is the line $y=-x$ in the complex plane?Or would we again say this is the line analogous to $y=-x$? $\endgroup$ – 68e1515 Jan 22 at 2:39
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    $\begingroup$ You should have, after dividing by$2$, the equation $|z-i/2|=2$. This is of course the circle of radius $2$ centered at $i/2$. $\endgroup$ – Chris Custer Jan 22 at 2:48
  • $\begingroup$ @68e1515 you'd be mapping single points to circles in the complex plane. $\endgroup$ – user645636 Jan 22 at 2:50
  • $\begingroup$ @ChrisCuster thanks for catching that typo! $\endgroup$ – Saketh Malyala Jan 22 at 2:56

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