# Complex plane conundrum

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.

• The vertical distance is just $1$ not $i$ – Peter Foreman Apr 25 at 17:50
• How is i equal to one, isn't the vertical axis for complex numbers – user650025 Apr 25 at 17:51
• 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. – Peter Foreman Apr 25 at 17:51
• 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. – KCd Apr 25 at 18:46

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 $$\mathbb R\times\mathbb R$$.