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.

enter image description here

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.

  • 2
    $\begingroup$ Absolute value defined this way has very useful properties. For example $|z+w| \le |z|+|w|$ and $|zw| = |z|\;|w|$. $\endgroup$ – GEdgar Aug 9 '16 at 18:14
  • 1
    $\begingroup$ When we write $z = x + iy$, we have $x, y \in \mathbb{R}$, so $(x,y)$ would be the representation of $z$ in $\mathbb{R}^2$. This leads to the same notion of distance as in the Pythagorean calculation. $\endgroup$ – Shagnik Aug 9 '16 at 18:15
  • $\begingroup$ $\mathbb C$ is precisely the vector space $\mathbb R^2$ with a multiplication operation added to the algebraic structure. That should answer all your questions $\endgroup$ – zhw. Aug 9 '16 at 18:16
  • $\begingroup$ Don't we also define $|z|$ as $ |z|^2=z \bar z$; in which case you get the correct formula (plus instead of minus). $\endgroup$ – ITA Aug 9 '16 at 18:18
  • $\begingroup$ @GEdgar, Are you saying that the absolute value is defined this way on purpose? I assumed it would arise as a consequence as a result of the structure of $\mathbb{C}$? Are you then saying that mathematicians chose to define the absolute value in this way for the properties it would evoke, when they defined $\mathbb{C}$? $\endgroup$ – Perturbative Aug 9 '16 at 18:24

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:

enter image description here

  • $\begingroup$ '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 $(x_1,y_1)(x_2,y_2)=x_1x_2+y_1y_2$)' - What you just said there is exactly what I was thinking (or felt was being done with this definition of the absolute value), i.e. switching back and forth between $\mathbb{R^2}$ and $\mathbb{C}$, which was why I got confused in the first place. You hit the nail on the hammer there. $\endgroup$ – Perturbative Aug 9 '16 at 18:26
  • $\begingroup$ In fact the image too, that you linked from Rudin where it says 'A vector space $X$ has dimension $n$ $(dim X = n)$ if $X$ has a *basis, $\{u_1, ..., u_n\}$. This means that every $x \in X$ has a unique representation of the form : $x = \alpha _1 u_1 + ... + \alpha_n u_n$'*, is EXACTLY, what I was trying to get at when I said (perhaps slightly incorrectly) that $z$ could be described as a 2-tuple over $\mathbb{R}$, but only as a 1-tuple over $\mathbb{C}$. Essentially why $z = (x, y)$ in $\mathbb{R^2}$, but $z = x+ iy$ in $\mathbb{C}$. Couldn't have asked for a better answer! $\endgroup$ – Perturbative Aug 9 '16 at 18:36
  • $\begingroup$ Glad to help, but quote from Rudin of interest is the "Example" about balanced sets (the definition is at the top), it may not yet be clear to you why anyone would care about balanced sets... but it does illustrate that in a very real way $\mathbb{C}$ and $\mathbb{R^2}$ have different vector space structure. $\endgroup$ – mb- Aug 9 '16 at 18:40
  • $\begingroup$ The reason a line segment L is not a balanced set in $\mathbb{C}$(intuitively, this means you can multiply an element in the set by a scalar of norm less than one, and stay in the set) is because you can multiply by a complex number $\alpha=e^{i\theta_2}$, with $|\alpha|=1$ to $x=re^{i\theta} \in L$ and $\alpha x=re^{i(\theta_1+\theta_2)}$ which need not be on the line, i.e multiplication by scalar field allows for rotations of elements about origin, this is absolutely not the case for $\mathbb{R}^2$! $\endgroup$ – mb- Aug 9 '16 at 18:41

Complex numbers $z=x+i*y$ are viewed as "a 2-tuple over $\mathbb{R}$" as you put but you made a mistake there which carried throughout your post.

You say $z=(x,iy)$ but this is not a tuple over $\mathbb{R}$ because $i$ is not a real number. The correct way would be to say that $z=x+iy=(x,y)$.

To answer your bolded question, why $i$ is dropped in the definition of $|z|$, is because $y$ is the length in the "imaginary" direction. If we look at the nice image that you posted, think about those 2 vectors in the image as vectors from the origin to the point $(x,y)=x+iy=z$ and to the point $(x,-y)=x-iy =\bar{z}$. Then the magnitude of a vector from the point $(0,0)$ to the point $(x,y)$ is defined by the Pythagorean Theorem as you mentioned, which would be $\sqrt{x^2+y^2}=|z|$.

  • $\begingroup$ So essentially $y$ is the Real length in the Imaginary Direction? And it's defined this way on purpose correct? $\endgroup$ – Perturbative Aug 9 '16 at 18:18
  • $\begingroup$ If you want to view $\mathbb{C}$ as an isomorphism to $\mathbb{R}^2$ we look at a complex number $z=x+iy=(x,y)$ as an ordered pair where $x$ is the real component of $z$ and $y$ is the imaginary component of $z$. It is defined in this way because we can see now that a real number is a special case of a more general complex number. If we take $y=0$ in $z=x+iy$ then we just have a real number. $\endgroup$ – Elliot Aug 9 '16 at 18:44

When $\mathbb{C}$ is identified with $\mathbb{R}^2$, we usually identify the number $1$ in $\mathbb{C}$ with the vector $(1, 0)$ in $\mathbb{R}^2$, and the number $i$ in $\mathbb{C}$ with the vector $(0, 1)$ in $\mathbb{R}^2$. Then, $x + yi = (x, y) = x(1, 0) + y(0, 1)$, so the coordinates of $x + yi$ w.r.t. the standard basis $\{(1, 0), (0, 1)\}$ of $\mathbb{R}^2$ are $(x, y)$ - and not $(x, iy)$, as you wrote, which would be impossible anyway because coordinates of a vector in $\mathbb{R}^2$ must be real.

  • $\begingroup$ (Comment reposted as an answer, as requested:) $\endgroup$ – Calum Gilhooley Aug 9 '16 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.