Both the real numbers and the complex numbers have a whole bunch of really nice properties: For reals, we have ordering, the intermediate value theorem, etc. Complex numbers are algebraically closed, and we have nice calculus results like the Cauchy–Goursat theorem, holomorphicity implies analyticity, etc. These results are to be contrasted with the case of e.g. higher dimensional spaces like $\mathbb{R}^n$ or other structures like the quaternions.

My feeling is that all the nice properties of the reals can be traced to completeness and ordering. However, I don't have a feeling for why the complex numbers have such miraculous analytical properties. Since $\mathbb{C}$ is defined essentially as the algebraic closure of $\mathbb{R}$, I might naively suspect that closedness is the crucial property, but I don't see how that manifests itself in proving theorems like Cauchy-Goursat. Perhaps Cauchy-Goursat is itself the essential property?

So my question is: Are there one or two fundamental properties of the complex numbers which beget all the other miracles of complex analysis? In other words, what makes complex analysis so different from real analysis or quaternionic analysis?


closed as too broad by Clayton, John Gowers, Adam Hughes, Shailesh, Namaste Dec 18 '16 at 0:59

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    $\begingroup$ because they have an imaginary part to them? $\endgroup$ – Jorge Fernández Hidalgo Dec 17 '16 at 20:04
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    $\begingroup$ See here for a related question. $\endgroup$ – Dietrich Burde Dec 17 '16 at 20:10
  • $\begingroup$ It's worth noting that the algebraic closure of $\mathbb C$ is often proved using Liouville's theorem (though it can certainly be proved with much less) - so the 'nice' analytic properties of $\mathbb C$ often come before the algebraic properties. $\endgroup$ – John Gowers Dec 17 '16 at 20:16
  • $\begingroup$ Multiplication adds a constraint that forces locally differentiable functions to be locally analytic. That is pretty amazing to me. $\endgroup$ – copper.hat Dec 17 '16 at 20:33
  • $\begingroup$ So your question is essentially why complex analysis (theory of holomorphic functions $\mathbb{C} \to \mathbb{C}$) is so "miraculous" compared to analysis in other spaces ? Maybe a proof that "differentiable $\implies$ analytic" is true only in (dense subspaces of) $\mathbb{C}$ is what you want ? The essential property is that if $f(z)$ is differentiable on $U$ and $\gamma \simeq \gamma'$ homotopically on $U$, then $\int_\gamma f(z)dz =\int_{\gamma'} f(z)dz$, plus the homotopy groups of open sets $U \subset \mathbb{C}$ $\endgroup$ – reuns Dec 17 '16 at 21:13

To me,the main thing that makes working in complex spaces-and it's very clear if you work strictly in the complex plane-is that since each complex number is determined by 2 components instead of one,then each corresponds to a point in a plane rather then a point on a line. As a result, mappings between complex numbers in the plane result in isometries and similarities that preserve oriented angles in the plane. This means we can use the tools of calculus in the complex plane to study classical geometry transformations and vice versa-something you can't do with real numbers.

For example, a good way to think of the derivative in the complex plane as a sequence of "infinitesimal" rotations of a tangent line to a circle centered at a point in the Argand plane-whereas the sequence of rotated tangent lines converges to the point by contracting in length along increasingly smaller subcircles. Also, most of the standard transformational geometry of the Euclidean plane has very elegant reformulations in terms of the standard analytic functions of the plane, such as the complex exponential in plane polar coordinates.

For this and much more on the geometric aspects of basic complex analysis, I can't recommend a better source then Tristan Needham's Visual Complex Analysis. It's a beautiful and deep text that everyone should read when studying complex analysis to really "get" why complex analysis is different from it's real counterpart.

  • $\begingroup$ 2D vectors also have 2 components. $\endgroup$ – Wood Dec 17 '16 at 21:28
  • $\begingroup$ @Wood Yes,but 2D vectors in an arbitrary vector space don't usually form a topological field. This is what allows us to manipulate their geometric properties in ways you can't do with ordinary 2D vectors.THAT was my point. $\endgroup$ – Mathemagician1234 Dec 18 '16 at 0:57

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