absolute valuecomplex-analysisdefinition
The definition I see everywhere for the absolute value of a complex number is:
Let $z=x+iy$ then $|z| := \sqrt{x^2+y^2}$.
But the square root operation is multivalued in complex analysis. So while this does show that the absolute value is a real number, it does not in fact indicate which square root we want. Of course it should be the nonnegative one, but what is the best way to express that? Would I just need to note "$|z| \ge 0$" in the definition? That's a little less than satisfying.
To Henry's comment here is the drawing:
The standard inner product of two complex numbers $z_1,z_2 \in \mathbb{C}$ is defined to be $z_1\bar{z_2}$, so the norm it induces will be $||z_1||:=\sqrt{z_1z_1}=\sqrt{z_1\bar{z_1}}=\sqrt{(a+bi)(a-bi)}=\sqrt{a^2+b^2}$. As a vector space over $\mathbb{R}$ (which it is almost never considered to be in practice), $\mathbb{C}$ has dimension two and thus is isomorphic to $\mathbb{R}^2$, the standard basis might be chosen to be $\{(1,0), (0,i)\}$ where we represent a number $z=a+bi$. The canonical isomorphic then sends $(1,0) \to (1,0)$ and $(0,i) \to (0,1)$, thus if you take a vector in $\mathbb{C}$, say $z=a+bi$, and "transport" it to $\mathbb{R^2}$ to investigate its norm in that inner product space (where the inner product is defined to be $(x_1,y_1)(x_2,y_2)=x_1x_2+y_1y_2$), we have $z=a+bi=a(1,0)+b(0,i)$, apply the canonical isomorphism and we get $a+bi$ is transported to $(a,b)$, (and not (a,bi)!), so the inner product here is different, there being no conjugation in the second coordinate, but that's fine because we're dropped the $i$ anyways by moving to $\mathbb{R^2}$. Pythagoras is fine.
One further comment: viewing $\mathbb{C}$ as a two-dimensional vector space over $\mathbb{R}$ is convenient merely for visualizing the complex plane, not for investigating its properties as a vector space, consider the following excerpt from Rudin's Functional Analysis:
Best Answer
We usually take it as a convention that $\sqrt{x} \geq 0$, but if you wanted to make it explicit, you could say something like "$|z| : = \sqrt{x^2 + y^2}$, where in the definition we adopt the convention that $\sqrt{r} \geq 0$."