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.

  1. 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 $s\rightarrow\infty$.

  1. 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.

  • $\begingroup$ Tag suggestions/edits would be much appreciated. $\endgroup$
    – OJFord
    Jun 2, 2015 at 12:38
  • $\begingroup$ Just a thought: Give me any point $z$ in the upper half complex plane and I can tell you that the semi-circular contour of radius $|z|+1$ encloses it. Rectangular regions can be constructed similarly. $\endgroup$
    – pshmath0
    Jun 2, 2015 at 12:42
  • $\begingroup$ @pbs Ah, I think you're onto it there. I suppose the shape doesn't matter so long as the plane is an open set of points. $\endgroup$
    – OJFord
    Jun 2, 2015 at 12:49
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    $\begingroup$ @Ollie: I sneaked in another link as an afterthought, but you'd already responded. :) Anyway, the linked Wikipedia pages (particularly the pages on the Riemann sphere and stereographic projection) will almost surely answer your question. The point is, the upper half-plane "is" a hemisphere, regardless of whether you use disks or rectangles or halves of regular dodecagons or irregular blobs to approximate (so long as your approximating sets "eventually contain every compact subset"). $\endgroup$ Jun 2, 2015 at 14:27
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    $\begingroup$ @OllieFord: Perhaps just "stereographic projection". This image is "the standard identification" of a sphere with the Riemann sphere. This image is evocative in quite a different way. I'll type up some overview details.... $\endgroup$ Jun 2, 2015 at 14:41

3 Answers 3


You could similarly take a square with $z=0$ in the middle of one of its sides, then let the square's sides tend to infinity. It doesn't matter really, as long as each point in the half-plane is contained in your growing square eventually. Often a semi-circle is the nicest choice because many complex objects are radial. For example, it's easier to estimate polynomials, something like $|z^n|$, on circles $|z| = r$ (there $|z^n| \leq r^n$) than on squares.


$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$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.

Stereographic projection from the north pole

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.

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    $\begingroup$ Thank you! This is exactly the sort of answer I hoped for, and clearly explained. Thanks for taking the time! $\endgroup$
    – OJFord
    Jun 2, 2015 at 17:29

When drawing some Nyquist plots, as the frequency increases, we know that the contour goes up to $0+\infty j$ on the imaginary axis. This point in the complex plane has a magnitude of $\infty$ and a phase of $\frac{\pi}{2}$. Then as the frequency increases even more, the magnitude of the current point on the Nyquist contour stays at $\infty$, but the phase changes from $\frac{\pi}{2}$ to $-\frac{\pi}{2}$.

A nice way to draw this is just to draw a semicircle with infinite radius. It makes sense because the radius is staying constant but the phase is changing, which leads to a circular arc. It doesn't really matter for most engineering purposes though, you just need to know that as you move along that contour, the phase is changing monotonically and the magnitude is constant at $\infty$.

  • $\begingroup$ It's not clear to me that anything you've said requires it to be defined as a semi-circle? Of course, it leads to a nice exponential model, but my question is really about the validity in the limit of - what appears to be - a polar construction on a rectangular grid. $\endgroup$
    – OJFord
    Jun 2, 2015 at 12:56
  • $\begingroup$ I was only providing an argument to show that a semi-circle is the most intuitive shape. I should have added that, as far as I am aware, the shape is not rigorously defined, it just needs to satisfy the properties I described above. Also, more importantly, why do you think the complex plane is a rectangular grid? You could just as easily and validly define a complex number in terms of its magnitude and phase, leading to a polar grid. I guess there must be an infinite number of alternate but valid ways to represent complex numbers, leading to different types of "grids". $\endgroup$
    – texasflood
    Jun 2, 2015 at 13:01
  • $\begingroup$ Also, a semi-square would not satisfy the property of constant magnitude as the phase changes. Only a semicircle would satisfy that. Depending on what you are plotting, constant magnitude for this section of the contour may be required $\endgroup$
    – texasflood
    Jun 2, 2015 at 13:03
  • $\begingroup$ Rectangular in the context of (among others) $z=x+iy$; at least common in my other example of Jordan's lemma. The shape, semi-circle or otherwise, of the contour would surely vary between coordinate systems - so we must surely have some concrete grid in mind? $\endgroup$
    – OJFord
    Jun 2, 2015 at 13:09
  • $\begingroup$ You could also say $z=\sqrt{x^2+y^2}e^{i\tan(y/x)}$, now it's polar. Yes, in general the shape would vary between coordinate systems, but it would be the same for cartesian and polar coordinates. For both cartesian and polar coordinates, a semicircle is a valid choice, and may be the only valid choice, depending on the problem. $\endgroup$
    – texasflood
    Jun 2, 2015 at 13:17

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