I was wondering whether there is a more common or 'best' way to define/think about the set of all complex number in complex analysis? The first way I can think of is:

$\Bbb{C} \triangleq \Bbb{R} \times \Bbb{I} = \{(a,b) \mid a\in\Bbb{R}, b\in\Bbb{I}\}$

Or is it more useful, down the road, to think of $\Bbb{C}$ as the set of all vector sums:

$\Bbb{C} \triangleq \{\vec{a}+\vec{b} \mid \vec{a}\in\Bbb{R}^1, \vec{b}\in\Bbb{I}^1\}$

and just remembering that you cannot add members/1-D vectors of $\Bbb{R}$ to members/1-D vectors of $\Bbb{I}$.

I guess another question would be how do you guys think about them?

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    $\begingroup$ What is $\;\Bbb I\;$ ?? $\endgroup$ – DonAntonio May 6 '16 at 21:02
  • $\begingroup$ the set of all imaginary numbers $\endgroup$ – Ben Granger May 6 '16 at 21:04
  • $\begingroup$ Like, say, $\;i\;$ or $\;-1.5 i\;$ ? None of the two ways you wrote in your question is known to me, at least ,as a standard way to express complex numbers. For me, these are pairs $\;a+bi\;$ , with $\;a,b\in\Bbb R\;$ and two well defined, nice operations: addition and multiplication. $\endgroup$ – DonAntonio May 6 '16 at 21:06
  • $\begingroup$ I suppose $\Bbb{I} = \{ai \mid a\in\Bbb{R}\}$ $\endgroup$ – Ben Granger May 6 '16 at 21:09
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    $\begingroup$ $\Bbb{C} \triangleq \Bbb{R} \times \Bbb{R} = \{(a,b) \mid a,b\in\Bbb{R}\}$. The set $\mathbb I$ is of little use. $\endgroup$ – Yves Daoust May 6 '16 at 21:09

For me the easiest, down-to-Earth way to see, at the beginning at least, the complex numbers is identifying $\;\Bbb C\sim\Bbb R^2\;$ , and definining here the well known complex addition and multiplication on the pairs:

$$(a,b)+(c,d):=(a+c, b+d)\;,\;\;(a,b)(c,d):=(ac-bd,ad+bc)$$

  • $\begingroup$ Does $\Bbb{C} \sim \Bbb{R}^2$ mean $\Bbb{C}$ is 'similar' to $\Bbb{R}^2$? $\endgroup$ – Ben Granger May 6 '16 at 21:11
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    $\begingroup$ Meaning $\Bbb{C}$ is the same as $\Bbb{R}^2$ except that it defines addition and multiplication differently? $\endgroup$ – Ben Granger May 6 '16 at 21:14
  • $\begingroup$ @BenGranger In factthey are isomorphic (as linear spaces over $\;\Bbb R\;$ , say), but this is unimportant: it is just "an identification" to begin with. To talk of abstract complex numbers is the same as talking to real, ordered pairs in the well kinonw real plane. $\endgroup$ – DonAntonio May 6 '16 at 21:17
  • $\begingroup$ @BenGranger The definitions are there, and in $\;\Bbb R^2\;$ there is not defined usually a multiplication. There is in this case. $\endgroup$ – DonAntonio May 6 '16 at 21:18
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    $\begingroup$ I always write something like $\Bbb C=(\Bbb R^2, +, \cdot_\Bbb C)$, and just define $\cdot_\Bbb C$ in the usual manner. $\endgroup$ – YoTengoUnLCD May 7 '16 at 0:21

I would say the most natural way from an algebraic standpoint is to view $\mathbb{C}$ as a field extension of $\mathbb{R}$. That is $\mathbb{C} \cong \mathbb{R}[X]/(X^2+1)$. This most clearly expresses that $\mathbb{C}$ is obtained from $\mathbb{R}$ by adjoining a square root of $-1$.

However while I feel this is the most natural way it does require knowledge of field extensions and it's a little less explicit than the other answers.

  • $\begingroup$ Hmmm yes I haven't quite gotten to field extensions yet, but thank you $\endgroup$ – Ben Granger May 6 '16 at 21:17
  • $\begingroup$ Understanding the most natural way to define something often requires being familiar with some algebra. Usually such definitions make the object both deep and crystal clear for me. +1 $\endgroup$ – lisyarus May 6 '16 at 21:19
  • $\begingroup$ I think this is pretty natural without much algebra, the complex numbers feel a lot like the real numbers with $i$ tossed in. $\endgroup$ – TokenToucan May 6 '16 at 21:24
  • $\begingroup$ So I could think about $\Bbb{C}$ as the field extension of $\Bbb{R}$ which satisfies the property $i = \sqrt{-1}$? $\endgroup$ – Ben Granger May 6 '16 at 21:28
  • $\begingroup$ @BenGranger If you know field extensions is then much simpler: $\;\Bbb C\;$ is the only non-trivial finite extension of the reals. $\endgroup$ – DonAntonio May 6 '16 at 21:29

To put my 2c in, I don't think there's a single 'best' way to think about the complex numbers (and I think this applies to thinking in general). Different situations demand different approaches, and being able to take 'shortcuts' by passing from one way of thinking about $\mathbb{C}$ to another can really help you see things intuitively. The polar and cartesian forms are the two most widely used ways to look at them. Moreover, knowing the different points of view can enhance your understanding of the core ideas behind the complex numbers (remember that any way of thinking about them is just a representation of these raw ideas).

Here are some ways I think about $\mathbb{C}$:

In more geometric problems, I think of addition as of translation, and of multiplication as of rotation followed by a scaling ('rotational homothety'). So for example, suppose you have a triangle $ABC$ and you build equilateral triangles outside of it on its sides, so that you end up with equilateral triangles with centers $X,Y,Z$. It's a classical problem (Napoleon's theorem) to show that $XYZ$ is equilateral. You can express equilateral-ness of a triangle $KLM$ using complex numbers: if you put the triangle anywhere in the complex plane so that $K,L,M$ correspond to complex numbers $k,l,m$ then, assuming $KLM$ is orienter counter-clockwise, it is equilateral iff $$ e^{i\pi/3}(l-k) = m-k $$ You can use that characterization (together with the fact that the center of an equilateral is the average of its vertices) to deduce the problem above by passing to the equivalent statement about complex numbers which is an easy calculation (good exercise).

In some algebraic problems, it is often useful to think of complex numbers as of performing some constructive/desructive interference of periodic stuff (the analogy is borrowed from physics, but there's a way we can make it accurate: think of the (discrete) Fourier transform, I wrote a bit about it here: http://amakelov.github.io/2016/01/28/Fourier-analysis-on-finite-abelian-groups.html). So for example if I want to compute $$S_0= \sum_{k=0}^{n/3}{n\choose 3k}$$ I have this intuition that I can use complex numbers to somehow pick out this periodic sequence (the arithmetic progression of numbers divisible by 3) from the larger thing I know, being $S=2^n$, the sum of all binomial coefficients. Just like we considered the expansion $$ (1+x)^n = \sum_{i=0}^n {n\choose i}x^i$$ to find $S$, we can plug in $x=\omega,\omega^2$ for $\omega$ a third root of unity and play around to get the result (try it out).

Here's a third way of thinking that I use when people start freaking out about this $\sqrt{-1}$ business not being real. You can interpret complex numbers as $2\times 2$ matrices, where $a+bi$ corresponds to the matrix $$ \begin{pmatrix} a & b\\ -b & a \end{pmatrix}$$ You can go ahead and verify that everything works as before in this new representation. The mysterious square root of $-1$ is now the matrix $$ \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$$ which promptly squares to minus the identity matrix. Why don't we just do it all that way? It is just much more convenient to think of complex numbers as single things obeying these special rules, and surely much faster to write them down. But I think convenience is not the only answer; thinking of something as a number just lets your brain do many useful things with it.

  • $\begingroup$ 'assuming KLM is orienter counter-clockwise, it is quadrilateral iff' Do you mean equilateral? $\endgroup$ – Ben Granger May 6 '16 at 22:03
  • $\begingroup$ yeah, sorry, it's fixed now. $\endgroup$ – amakelov May 6 '16 at 23:44
  • $\begingroup$ I just drew that out on the complex plane, that's a really cool way to prove equilateral-ness $\endgroup$ – Ben Granger May 6 '16 at 23:45

"I was wondering whether there is a more common or 'best' way to define/think about the set of all complex number in complex analysis." There really isn't. At different times you will draw on different aspects of $\mathbb{C}$. Sometimes you'll analyze it as a vector space. Sometimes you'll look at it as an algebraically closed field. Other times it will be an algebra. Later on they may be linear similarity transformations on $\mathbb{R}^{2}$. Maybe they'll be polar expressions of the plane (i.e. $z = r e^{i \theta}$, where $r \geq 0, \theta \in [0, 2 \pi)$).


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