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So I've been wondering about geometry on the complex plane. Points on this plane are denoted by $(a+bi)$, right? If we have the point $(1+i)$, the horizontal distance from the complex axis is $1$, and the vertical distance from the real axis is $i$. So the distance from the origin by Pythagorean theorem would be: $$ 1^2 + i^2$$ $$=1+(-1)$$ $$= 0?$$ But that would mean the point $(1+i)$ is on the origin which it it obviously not! Is there an explanation or is regular geometry like Pythagorean theorem not applicable to the complex plane.

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    $\begingroup$ The vertical distance is just $1$ not $i$ $\endgroup$ – Peter Foreman Apr 25 at 17:50
  • $\begingroup$ How is i equal to one, isn't the vertical axis for complex numbers $\endgroup$ – user650025 Apr 25 at 17:51
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    $\begingroup$ Yes, but distance is always real. We just use the complex plane to geometrically represent complex numbers as coordinates with real parts on the $x$ axis and imaginary parts on the $y$ axis. $\endgroup$ – Peter Foreman Apr 25 at 17:51
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    $\begingroup$ The tag is wrong. Despite what you may think, the term "complex geometry" is not meant to refer to this type of topic, but something more sophisticated. $\endgroup$ – KCd Apr 25 at 18:46
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As already mentioned in the comments, distance functions are typically real valued, not complex valued.

A distance function on a set $X$ is a map $X\times X\to \mathbb R$ that satisfies certain properties. Two of the properties include inequalities.

If you try to replace $\mathbb R$ with $\mathbb C$, this stops making sense because there is no ordering for $\mathbb C$ for those axioms to work with, at least, not one that is geometrically useful. Instead, $\mathbb C$ gets its metric from its underlying metric space $\mathbb R\times\mathbb R$.

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