Let's say we have an arbitrary complex number $z \in \mathbb{C}$ , $z = x+iy$
Then the absolute value (or magnitude/norm of $z$) is defined as follows.
$$|z| \stackrel{\text{def}}{=} \sqrt{x^2 + y^2}$$
But to me it seems a bit hand wavy as that is exactly the magnitude of a vector in $\mathbb{R^2}$ e.g $||\vec{a}|| = \sqrt{(a_1, a_2) \bullet (a_1, a_2)} = \sqrt{a_{1}^{2} + a_2^2}$
I asked a question earlier if it was possible to write a complex number as a vector in $\mathbb{R}^2$ : Writing Complex Numbers as a Vector in $\mathbb{R^2}$, and the answer was yes, as $\mathbb{C}$ is isomorphic to $\mathbb{R^2}$, however one has to be careful with how you choose to write it.
Am I correct in saying that you can describe a point in the complex plane as a 2-tuple $(x, iy)$ over $\mathbb{R^2}$ but only as a one tuple, a scalar $x+ iy$ over $\mathbb{C}$. i.e. $\mathbb{C}$ is one-dimensional with respect to itself, but two-dimensional with respect to $\mathbb{R}$. I ask this as there may be a misinterpretation on my part.
I've included this image here to illustrate my point.
To find the absolute value (the magnitude) of $|z|$, if $z = (x, iy)$ (a 2-tuple over $\mathbb{R}$, or a point in $\mathbb{R^2}$), then wouldn't the absolute value be, by the Theorem of Pythagoras (or via the square root of the dot-product with itself)
$$|z| \stackrel{\text{def}}{=} \sqrt{x^2 + i^2y^2} \implies |z| \stackrel{\text{def}}{=} \sqrt{x^2 – y^2}$$
Now this can't be right, so my question boils down to:
Why is the $i$ just dropped in the definition of the absolute value of $|z|$?
We certainly can't use the Theorem of Pythagoras over $\mathbb{C}$ as $\mathbb{C}$ is one-dimensional with respect to itself, so we must be using the Theorem of Pythagoras over $\mathbb{R^2}$, and in that case one of the basis vectors must be $(0, i)$, correct? (i.e it must contain the Imaginary Axis for $\mathbb{R^2} = \mathbb{C}$)
If I'm totally off the ball here, please tell me as it seems I'm having trouble making the connection between $\mathbb{R^2}$ and $\mathbb{C}$. I've included extra information in my question so that you can see where I'm coming from when I make the arguments I'm trying to make.
Best Answer
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: