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I was searching around for whether or not I could use the Euclidean distance with complex numbers because it seems to make sense but I was struggling to really know. I found someone say that it is true that we can use the euclidean distance because $\mathbb{C}$ is isomorphic to $\mathbb{R}^2$ and that just doesn't seem right to me at all, however when I research it I keep finding conflicting information.

I think I'm misunderstanding what isomorphic means here, because it seems to me that group operations aren't preserved between the two.

Blake
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  • Say what's true? This question is very unclear the way it's phrased. –  Jun 17 '17 at 02:26
  • The claim was that "It's true that you can use Eucliean Distance for Complex Numbers" and justified by "$\mathbb{C}$ is isomorphic to $\mathbb{R}^2$" – Blake Jun 17 '17 at 02:27
  • Yes, it's true. The complex number $a + bi$ corresponds to the point $(a,b)$ in the $xy$-plane. So to find the Euclidean distance between $z = a + bi$ and $w = x + yi$, just find the Euclidean distance between the points $(a,b)$ and $(x,y)$. –  Jun 17 '17 at 02:27
  • $\mathbb{C}$ is isomorhpic to $\mathbb{R}^2$, not $\mathbb{R}$, so your "hunch" is misguided. The fact that $\not\exists n\in\mathbb{R}:n^2=-1$ doesn't tell you anything. – user160738 Jun 17 '17 at 02:28
  • What you can't do is compare two imaginary numbers in the same way that you can compare real numbers, e.g., $1 < 2$. There's no corresponding notion of "order" for imaginary numbers. –  Jun 17 '17 at 02:28
  • So, then, is $\mathbb{C}$ isomorphic to $\mathbb{R}^2$ and how can I know that? – Blake Jun 17 '17 at 02:30
  • I gave you the isomorphism above without calling it an isomorphism. The point $a+bi$ in $\Bbb C$ corresponds to the point $(a,b)$ in $\Bbb R^2$. –  Jun 17 '17 at 02:31
  • Maybe you are thinking at $|z_1-z_2|^2=(x_1-x_2)^2+(y_1-y_2)^2$ where $z_k=x_k+ i y_k$ for $k=1,2$. – dxiv Jun 17 '17 at 02:32
  • My issue here is that maybe I misunderstand or misremember Isomorphism as preserving group operations? – Blake Jun 17 '17 at 02:34
  • @Blake See What's the difference between $\mathbb{R}^2$ and the complex plane?. – dxiv Jun 17 '17 at 02:38
  • It is only isomorphic to $\mathbb{R}^2$ if you define a suitable multiplication on $\mathbb{R}^2$. – copper.hat Jun 17 '17 at 02:40
  • However, you can use the distance $d(z,w) = |z-w|$ regardless of any isomorphism. – copper.hat Jun 17 '17 at 02:49
  • The first secret is the shortcut $z \overline{z} = \overline{z} z = |z|^2$ which boils down "by accident" to as if it was meant for the euclidean metric over $\mathbb{R^2}$ – user76568 Jun 17 '17 at 03:41

2 Answers2

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Isomorphic here is regarding $\mathbb{C}$ and $\mathbb{R^2}$ as both being $2$ dimensional vector spaces over the field $\mathbb{R}$.

Vector spaces after all. Or before..

Let's see you get a non zero real number from a linear combination of smiley faces and frowny faces. :D

user76568
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The function

$$f(a+ib)=(a,b)\in \mathbb{R}^2$$

is a (field) isomorphism, where $\mathbb{R}^2$ is usual group with pointwise addition and multiplication is defined as it should be:

$$(a,b)\cdot(c,d)=(ac-bd,ad+bc)$$

This gives you $f(z_1+z_2)=f(z_1)+f(z_2)$ and $f(z_1z_2)=f(z_1)f(z_2)$.

Since once can define Euclidean metric on $\mathbb{R}^2$ then you can traslate this definition back to $\mathbb{C}$ via isomorphism.

user160738
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