TITLE: Why do we assume the complex plane is curvey at infinity?
QUESTION [4 upvotes]: In at least a few areas (i.e., those that I have happened across) when we have a need to capture a half-plane we do so by taking a semi-circle of radius $r$ in that half, and taking the limit as $r\rightarrow\infty$.
For example, the positive-imaginary half-plane for Jordan's Lemma, or the positive-real half-plane to construct the Nyquist Contour.

What motivates this model?

It would seem to me that in a 'Cartesian-like' grid, we should have to take a semi-square of side length $s$ (that is, an $s, 2s$ sided rectangle) in the limit $s\rightarrow\infty$.

How do we know the semi-circle construction is equivalent? (Or, why am I wrong, and the rectangular construction doesn't work?)

I have a sort of sketchy idea from discussion in comments of @texasfloods' answer that I can't think how to formalise - that given the unit circle $x^2+y^2=r^2$ it seems intuitive that taking $\lim_{r\rightarrow\infty}$ is identical to $\lim_{\sqrt{x^2+y^2}\rightarrow\infty}$, the hypotenuse of the right triangle, and by a bit of hand-waving, "done".
Also from @pbs' comment below, it seems reasonable to say that for any open set $A$ of points enclosed by any 'contour shape' $\Gamma, \forall z\in A$ there exists a 'large-enough' construction of $\Gamma$ to enclose $z$.
I'm fairly convinced, but I'd still be interested in a more 'proper'/rigorous proof or explanation if anyone cares to offer one.

REPLY [3 votes]: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$Let $z = x + iy$ denote a general complex number expressed in terms of its real and imaginary parts, and let $(u, v, w)$ denote Cartesian coordinates on $\Reals^{3}$.
If $S^{2} \subset \Reals^{3}$ denotes the unit sphere centered at the origin, with "north pole" $N = (0, 0, 1)$, there is a simple, remarkable identification of the complex plane $\Cpx$ with $S^{2} \setminus\{N\}$, the complement of the north pole: For each point $(u, v, w) \neq N$ of the sphere, i.e., for each triple $(u, v, w)$ with $u^{2} + v^{2} + w^{2} = 1$ and $w < 1$, there is a unique line in $\Reals^{3}$ from $N$ to $(u, v, w)$. This line crosses the $(u, v)$-plane at a location $(x, y, 0) = \Pi(u, v, w)$. Conversely, every point of the $(u, v)$-plane arises (exactly once) in this way.

Using similar triangles, vector algebra, or comparable techniques, it's easy to discover algebraic formulas for this mapping by stereographic projection (from the north pole):
$$
(x, y) = \Pi(u, v, w) = \frac{(u, v)}{1 - w},\qquad
(u, v, w) = \Pi^{-1}(x, y) = \frac{(2x, 2y, x^{2} + y^{2} - 1)}{x^{2} + y^{2} + 1}.
$$
If we identify a complex number $z$ with $x + iy$, stereographic projection from the north pole identifies the complex plane with the complement of the north pole in the unit sphere $S^{2}$. The "missing" point $N = (0, 0, 1)$ may be viewed as $\infty$, "the point at infinity in $\Cpx$". The name is apt, since if $(z_{k})$ is a sequence of complex numbers whose magnitudes grow without bound, then the sequence $\bigl(\Pi^{-1}(z_{k})\bigr)$ in the sphere converges to $N$.
The mappings $\Pi$ and $\Pi^{-1}$ posses remarkable geometric properties:

Conformality: Each map preserves the angles at which curves meet. (There is a beautiful, elementary geometric proof which this post is a little too small to contain.)
Circle-preservation: The image under $\Pi$ of a circle $C$ in the sphere is either a circle or a line ("a circle of infinite radius") in $\Cpx$, depending on whether $C$ misses $N$ or passes through $N$. Conversely, if $C$ is a circle or a line in $\Cpx$, then the image $\Pi^{-1}(C)$ is a circle in the sphere.

Consequently, the stereographic image of an open hemisphere $H$ in $S^{2}$ is either the inside of an open disk (if the north pole is not in the closure of $H$), a half-plane (if $N$ lies on the boundary of $H$), or the outside of a disk (if $N$ lies in $H$). In particular:

The upper half plane $\{z : \Im z > 0\}$ corresponds to the open "eastern" hemisphere $\{(u, v, w) : v > 0\}$.
The right half plane $\{z : \Re z > 0\}$ corresponds to the open hemisphere $\{(u, v, w) : u > 0\}$.
The unit disk $\{z : |z| < 1\}$ corresponds to the open "southern" hemisphere $\{(u, v, w) : w < 0\}$.

(Exercise: What corresponds to the "northern" hemisphere?)

 Answer: The outside of the closed unit disk: $\{z : |z| > 1\}$.

The first of these (finally) addresses the question: The upper half-plane in $\Cpx$ is "essentially" an open hemisphere. If $(U_{k})$ is a sequence of nested sets, and if every compact (closed and bounded) subset of the upper half-plane is contained in some $U_{k}$, then the union of the $U_{k}$ is the upper half-plane, which corresponds to the "eastern" hemisphere.
It's not so much that the complex plane is circular near infinity, it's that in the "one-point compactification" of $\Cpx$, the inverse projection $\Pi^{-1}$ maps "large" boundary curves to "small" curves near the north pole, and in the limit, boundaries are squashed to a point.

Since the story has come this far, it's worth pointing out that the sphere admits a complex-analytic structure near $N$ if we define $\zeta = 1/z$ to be a complex coordinate, and if we define $\zeta = 0$ at the north pole. (Yes, that means "$1/0 = \infty$" and "$1/\infty = 0$" in this specific context.) Contrary to one's possible first impression, however, the coordinate $\zeta$ is not induced by stereographic projection from the south pole $S = (0, 0, -1)$, but by projection from the south pole followed by complex conjugation, a.k.a., reflection across the real axis.
Inverse projection from the north pole followed by projection from the south pole is a conformal mapping (or, to be picky, an anti-conformal mapping) on $\Cpx \setminus\{0\}$ known as inversion in the unit circle. In polar coordinates, $(r, \theta) \mapsto (\frac{1}{r}, \theta)$. On the sphere, inversion in the unit circle corresponds to reflection in the $(u, v)$-plane. That is, the points $(u, v, w)$ and $(u, v, -w)$ map to points (lying on a ray through the origin) whose magnitudes are mutually reciprocal.