Ive always thought that infinity isn't really a number. It is just an idea - a name we attach to something that grows without bound.

So in real analysis, when the terms of a sequence or partial sums of a sequence (series) keep increasing without an upper bound, we say the sequence or the series goes to infinity. Negative infinity is the same idea, but with a minus sign, that is negative terms, which keep decreasing without any lower bound go to $-\infty$. And this "infinity" object is bigger than any number you could possibly think of (because it isn't a number in itself). All fine.

But attached with this idea of bigger than everything else, is the notion of big or small, i.e. order. However for complex numbers, there is no total order. We just can't compare any 2 given complex numbers and say which is "bigger". How, then is infinity thought of in complex analysis? It can't be an element that is bigger than all other elements, because "bigger" doesn't make any sense.

And since there is only one infinity in the complex plane, unlike $+\infty$ and $-\infty$ in $\mathbb{R}$, does that mean that the complex infinity is more of a scalar (with only a modulus and no direction or argument) than a vector (like we can associate a vector with all other complex numbers)?

I know I sound very confused. I am. Please shed some light. Thanks.

  • $\begingroup$ Infinity is simply the set of nonfinite. Anything that does not converge to finite nor gets into a bounded loop ends up in infinity. $\endgroup$
    – mick
    Commented Sep 10, 2017 at 14:43
  • $\begingroup$ Yes I think of it more as a scalar. $\endgroup$
    – mick
    Commented Sep 10, 2017 at 14:44
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    $\begingroup$ You actually seem to have figured it out on your own. Complex infinity is a concept relating to what happens when the modulus grows without bound while the direction is not determined. $\endgroup$
    – David K
    Commented Sep 10, 2017 at 14:52
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    $\begingroup$ In complex analysis we often need the idea of "continuous at $\infty$". For example $\frac{1}{z}$ is continuous at $\infty$. And $e^{-z}$ is continuous at the $\infty$ of $\Re(z) \ge \epsilon > 0$. So there are many different infinities, and which one you need depends on the context. $\endgroup$
    – reuns
    Commented Sep 10, 2017 at 15:21
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    $\begingroup$ @mick Expressions like "the set of nonfinite" are precisely why one needs to be careful with the definition of sets. $\endgroup$
    – Kimball
    Commented Sep 10, 2017 at 17:03

3 Answers 3


$\infty$ is the compactification point of the complex plane. In this context, $\infty$ is very real: it is the north pole of the Riemann sphere. $0$ is the south pole.

In fact, $0$ and $\infty$ are closely related:

  • they both only have a modulus and no direction or argument.

  • the function $z \mapsto \dfrac 1z$ exchanges them.

  • they $0$ and $\infty$ are absorbing elements for multiplication: $0 \cdot z = 0$, $\infty \cdot z = \infty$, for $z \ne 0, \infty$.

  • $\begingroup$ the comment to the answer above, holds also for yours.The limit of a sequence of complex numbers does it not depend on how (with which phase) the "compactification point" is reached ? $\endgroup$
    – G Cab
    Commented Sep 10, 2017 at 15:57
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    $\begingroup$ While some (or many) may prefer concise rigorous answers such as this, it presupposes understanding by the asker of concepts the asker likely doesn't understand well, which is why they are asking the question in the first place. So I'm inclined to call it "rigorous, concise, and unhelpful" and not worthy of an upvote (and not nearly bad enough for a downvote). $\endgroup$ Commented Sep 11, 2017 at 0:14
  • $\begingroup$ Hm, the function $z \mapsto -1/z$ also exchanges them, right? What's so special about that one? $\endgroup$
    – user541686
    Commented Sep 11, 2017 at 0:16
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    $\begingroup$ @Mehrdad: The same thing that makes $i$ "special" in comparison to $-i$: It has fewer symbols so it looks tidier. $\endgroup$
    – Kevin
    Commented Sep 11, 2017 at 4:32
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    $\begingroup$ There is no such thing as "the compactification of $\mathbb{C}$". The Riemann sphere is a common one, but there are many others (a disk, a projective plane...). $\endgroup$ Commented Sep 11, 2017 at 7:50

In the complex plane, it is useful to first think of how we change the meaning of the term "bounded". In the reals, "bounded by $M$" means "$|x| < M$". In the complexes, "bounded by $M$" looks like it means the same thing, "$|z| < M$", but $|\cdot|$ has changed from real magnitude to complex magnitude.

We might say that a sequence in the reals is unbounded if for any bound $M$, the sequence is eventually permanently made up of terms that are greater in magnitude than $M$. (We might, but we shouldn't. We should say "unbounded" means that for any $M$, the absolute values of the sequence at least once exceed $M$, so that no bound is large enough to dominate the entire sequence. However, it is useful to think about sequences that are monotonic to fix one's ideas before thinking about general sequences.) Note that we haven't said if they're increasing towards $+\infty$ or decreasing towards $-\infty$. We've said they're unbounded in magnitude without picking a direction, so the order of the reals isn't entering into this idea. We can do the same thing in the complexes -- if a sequence eventually permanently exceeds any given bound, then the sequence is unbounded in magnitude. If we were going to be technically sound, we would say of sequences of points, $x_1, x_2, \dots$ and $z_1, z_2, \dots$, "$|x_n|$ goes to infinity" or "$|z_n|$ goes to infinity." Again, we are ignoring the order and only attending to the magnitude of the numbers. If this is where we stop, there is only one infinity. Under stereographic projection of the plane onto the unit sphere, that infinity appears as the north pole. See "Riemann sphere", where the circles of larger and larger magnitude are eventually drawn up (like a bag) to a single point, labelled "$\infty$". When we treat infinity in the complexes this way, it is easier to talk about a "neighborhood of infinity" -- it's a disk of points around the north pole in the projection. When we un-project, that's the set of points outside a disk centered at $0$. One reason this way of thinking of infinity is useful is that now every point on the sphere "is the same" -- any little neighborhood you pick around one point can be slid along the sphere to another point, even the point at infinity. So computations that previously only made sense at finite points can be made sensical at $\infty$.

If we do pay attention to the order, we can be a little more precise and, in the reals, say "$x_n$ goes to $+\infty$" if for any bound $M>0$, the sequence eventually permanently above that bound and say "$x_n$ goes to $-\infty$" if for any bound $M<0$, the sequence is eventually permanently below that bound. If we pretend that these sequences are sequences on the real line in the complexes, then we see that in the one case, we are going in the direction $\mathrm{e}^{\mathrm{i} 0} = 1$ and in the direction $\mathrm{e}^{\mathrm{i} \pi} = -1$ and notice that we said we were going to $1 \cdot \infty$ and $-1 \cdot \infty$, if we allow ourselves (only) the tiniest amount of algebra with infinities. But we can take a sequence $x_n$ going to $+\infty$ and make a new sequence $z_n = \mathrm{e}^{\mathrm{i} \theta} x_n$, which is a sequence whose magnitude increases without bound and whose direction is along the ray starting at $0$ and passing through $\mathrm{e}^{\mathrm{i} \theta}$. Could we not generalize our idea of infinity to allow directed infinities? "$z_n$ goes to $\mathrm{e}^{\mathrm{i} \theta} \infty$". This idea expresses that for any bound $M>0$ and any tiny error in the argument $\delta \theta>0$, the sequence eventually permanently has $|z_n| > M$ and $\arg(z_n) \in (\theta - \delta \theta, \theta + \delta \theta)$. Now we can more precisely describe sequences that run away from the origin asymptotically approaching a ray from the origin -- a form of going to a signed infinity where the sign can be any point on the complex unit circle, just as it was for signed infinities in the reals (where there were only two points on the unit circle, $\pm 1$).

We can generalize further, and this process of deciding what to do to sequences that try to "run off the edge" (whatever that means for infinite spaces having no concrete edges) leads to points at infinity. There are many choices. The Alexandroff one-point compactification makes the choice that there is only one $\infty$, like our magnitude choice above, and chooses to treat all eventually permanently exceeding any given bound sequences as all converging to the "one point", $\infty$. This gives variations of the Riemann sphere for whatever space you do it to. If we follow the directed infinity path, then we get various projective spaces, but at least we can keep track of the direction in which we ran away to infinity. There are additional ways to collect sequences running away to infinity into classes and assign a "point at infinity" for the members of that class, leading to various compactification schemes, but this is probably more generalization than you need for your current thoughts on infinity.

  • 2
    $\begingroup$ Thanks, this is a more complete and clear answer, finally allowing to go over the Riemann Sphere cited as a totem. $\endgroup$
    – G Cab
    Commented Sep 10, 2017 at 16:46
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    $\begingroup$ Very nice answer (+1) $\endgroup$
    – tired
    Commented Sep 10, 2017 at 21:39
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    $\begingroup$ "Unbounded" does not necessarily means "goes to infinity" – e.g. $a_n = n·( (-1)^n + 1)$ is an unbounded sequence, but does not have any limit at all (in any compactification). $\endgroup$ Commented Sep 10, 2017 at 22:06
  • $\begingroup$ @PaŭloEbermann : I'm trying to figure out how your observation applies. Every use of "goes" in this answer is contingent on a particular quantified predicate I am hiding behind "eventually permanently" past a bound. Your sequence does not satisfy that quantified predicate nor the phrase "eventually permanently" past a bound. $\endgroup$ Commented Sep 11, 2017 at 1:21
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    $\begingroup$ "Unbounded" just means it has infinitely many arbitrarily large terms. Not that it is eventually permanently large. I understand what you're trying to say, and it is a well-written answer, but "unbounded" is the wrong word to use. So I agree with @PaŭloEbermann $\endgroup$
    – Arthur
    Commented Sep 11, 2017 at 6:03

There is no order in $\mathbb C$. Right! But if $z\in\mathbb C$, $|z|$ can be big or it can be close to $0$. So, take a sequence $(z_n)_{n\in\mathbb N}$ of complex numbers. Suppose that $\lim_{n\to\infty}|z_n|=+\infty$. What this means is that the (real) numbers $|z_n|$ are arbitrarily large. So, you can state this saying that $\lim_{n\to\infty}z_n=\infty$ (not $+\infty$).

Another example. Let $f(z)=\frac 1z$. If $z$ is very far from $0$ (that is, if $|z|$ is very large), then $f(z)$ is very close to $0$. We express this saying that $\lim_{z\to\infty}f(z)=0$. The formal definition is$$(\forall\varepsilon>0)(\exists R>0):|z|>R\implies\bigl|f(z)\bigr|<\varepsilon.$$

Note that in $\mathbb C$ we only have one $\infty$; we don't have a $+\infty$ and a $-\infty$.

For more about this, read about the Riemann sphere.

  • $\begingroup$ ok, but what about , e.g., the sequence $z_n=(n+in)$ ? $\endgroup$
    – G Cab
    Commented Sep 10, 2017 at 15:17
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    $\begingroup$ @GCab $\lim_{n\to\infty}z_n=\infty$, because $(\forall n\in\mathbb{N}):|n+in|=n\sqrt2$ and $\lim_{n\to\infty}n\sqrt2=+\infty$. $\endgroup$ Commented Sep 10, 2017 at 15:18
  • $\begingroup$ for the modulus of $z$, that' s clear. But for $z$ as complex number can't we say that $lim_{n\to\infty}z_n=(1+i)\infty$ ? $\endgroup$
    – G Cab
    Commented Sep 10, 2017 at 15:23
  • $\begingroup$ @GCab No. As I wrote in my answer, in $\mathbb C$ there is only one $\infty$. $\endgroup$ Commented Sep 10, 2017 at 15:32
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    $\begingroup$ @GCab : Yes. See directed infinity. $\endgroup$ Commented Sep 10, 2017 at 15:36

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