What is the meaning of $z \to \infty $ when working with complex numbers?

Question

Let $z=x+iy$. If I am evaluating $f(z)$ as $z$ tends to infinity, I should assume this means $x$ tends to infinity and $y$ tends to infinity?

I think this is the case but wanted to confirm. I realized this when checking the solution to a problem asking for the limit of $f(z)=e^z$ as $z$ tends to infinity. At first I simply thought infinity was the answer as I was "thinking" of $z$ as infinity $+ 0i$, but I see that it's periodic when thinking of $x$ and $y$ both tending to infinity. Obviously if it said the limit as $z$ tends to $5$, the answer would be $e^5$, but I suppose when thinking of infinity we consider both the real and imaginary parts to tend to infinity?

Answer 1

It's not necessary for both $x$ and $y$ to tend to infinity, just that one or the other tends to infinity. To understand what's going on we should consider the Riemann sphere, which is the stereographic projection of the complex plane.

To visualize this we take the unit sphere so that the equator intersects the $x,y$-plane which we will identify with the complex plane. We not take a line from the point $(0,0,1)$ on the sphere and connect it to some point on the plane $(x,y,0)$. This line will intersect the sphere at exactly one point with points inside the unit circle being mapped to the lower hemisphere and points outside the unit circle being mapped to the top hemisphere. This shows there is a bijection between points on the plane and points on the sphere without the point $(0,0,1)$.

However if I keep traveling away from the origin indefinitely I will be approaching the point $(0,0,1)$. The direction doesn't matter, we just need the length of $(x,y,0)$ to grow indefinitely. If you consider lines in the complex plane they will form great circles on the sphere that go through the point $(0,0,1)$ so even if you travel along the $x$ or $y$ axis you'll approach $(0,0,1)$. This is what motivates us to define $\infty = (0,0,1)$ in this fashion.

One interesting fact about the Riemann sphere is that the map $z \rightarrow \frac 1z$ inverts the sphere, mapping $\infty$ to $0$ and vice versa. This gives us an easy way to tell if something converges to $\infty$ since the reciprocal will converge to $0$. Since we don't have to appeal to the geometry to discuss the reciprocal map this gives us an analytic means to define convergence to infinity saying $f(z)$ converges to $\infty$ if $\frac 1{f(z)}$ converges to $0$. The geometric picture is useful for intuition and this second method is useful for calculation.

Answer 2

If I am evaluating $f(z)$ as $z$ tends to infinity, I should assume this means $x$ tends to infinity and $y$ tends to infinity?

This is not true.

When we say $z\to\infty$, it means $|z|\to\infty$. For example, you may choose $y=0\land x\to\infty$, which means $z$ approaches to infinity along the positive real axis. Or you may choose $y=0\land x\to-\infty$, which means $z$ approaches to infinity along the negative real axis. Also, you may choose $x=0\land y\to\infty$, which means $z$ approaches to infinity along the positive imaginary axis, etc.

Note that above examples are approaching to infinity along straight lines, but it is not necessary to approach to infinity along straight lines. You may choose any curves (we call it "path"), as long as $|z|\to\infty$. Hope this solves your confusions.

Answer 3

In real analysis, it is possible to make sense of the statement $\lim_{x\to \infty}f(x)=l$ in the context of real numbers alone, and without having to define what the symbol "$\infty$" means. Specifically, we can think of $\lim_{x\to \infty}f(x)=l$ as a mere notional shorthand for the longer statement $$ (\forall\varepsilon>0)(\exists M)(\forall x)(x>M\implies |f(x)-l|<\varepsilon) \, . $$ (Here, I'm assuming that $l$ is a real number. Exercise: how would you define the statement $\lim_{x\to\infty}f(x)=\infty$?)

If we tried to naively extend this definition to complex analysis, we do end up in trouble: it makes no sense to define $\lim_{z\to\infty}f(z)=l$ as $$ (\forall\varepsilon>0)(\exists M)(\forall z)(z>M\implies |f(z)-l|<\varepsilon) \, . $$ The issue is that there is no natural way of defining what $z>M$ means in the context of the complex numbers.

However, this issue disappears if we instead intuitvely think of $z\to\infty$ as meaning "the absolute value of $z$ becomes arbitrarily large", since the absolute-value function is real-valued, and there is a natural ordering of the real numbers.

This leads to the following definition: $\lim_{z\to\infty}f(z)=l$ is a shorthand for $$ (\forall\varepsilon>0)(\exists M\in\mathbb R)(\forall z)(|z|>M\implies|f(z)-l|<\varepsilon) \, . $$ As mentioned in the comments, we can also equivalently define $\lim_{z\to\infty}f(z)=l$ as meaning $\lim_{w\to 0}f(1/w)=l$, but I feel that the above definition is more natural.

I should add that while statements about limits to infinity can always be interpreted as a shorthand for statements about real/complex numbers, it is sometimes fruitful to define $\infty$ as a specific object. This allows us to see the different types of limits as instances of a single overarching concept (and there are many other uses). In real analysis, we have the affinely extended real line, and the projectively extended real line. The "$\infty$" of the former behaves differently to the "$\infty$" of the latter, and so it is important to be precise about which specific notion of infinity one is using. In complex analysis, there is the Riemann sphere. All of these infinities behave differently to regular numbers. Tread carefully.