How to compute limit to infinity of an exponential in complex analysis?

Question

My question originates from trying to find the following limit:

$$ \lim_{z\to\infty} e^{-z^2} $$

Disclaimer: before getting into details, I want to point out that English isn't my first language, so I apologise for grammar mistakes, and that I'm a Physics student who is fairly new to this branch of mathematics.

I've found countless questions on this site dealing with much tougher exercises, but nothing that takes into account the very basic details of each step in calculating this kind of limits. On the other hand, my textbooks (LaTeX notes from two different professors, where I found the exercise, and John B. Conway's second edition of Functions of One Complex Variable I) all show a very meticulous approach to the subject (not the limit - the exercise I'm dealing with has no solution anywhere in my notes), in which I can't seem to get a solid understanding of the whole process. I just get lost in a lot of computation and theory.

The limit written above comes right after a section on singularities, as an exercise on singularities going to infinity. First of all, I can't figure out whether $\infty$ is an actual singularity for $e^{-z^2}$, nor if this is required to know. My thought process started with writing the Taylor series expansion, which I reckon should be in fact its Laurent series expansion:

$$ \lim_{z\to\infty} e^{-z^2} = \lim_{z\to\infty}\sum_{k=0}^{+\infty}\frac{\left( -z^2 \right)^k}{k!} = \lim_{z\to\infty}\sum_{k=0}^{+\infty}\frac{\left( -1 \right)^k}{k!} z^{2k} $$

At this point, I find it ambiguous to require $z\to\infty$, since the professor is prone to typos; I thought that maybe he intended to write $|z|\to\infty$ instead. I'm not sure, maybe the limit makes sense in both cases. Anyways, I had two ideas:

1. case: $|z|\to\infty$. I just write explicitly: $z=\rho e^{i\theta}$ with $\rho\to\infty$. I guess the limit doesn't exist, since its absolute value goes to infinity but there's no condition on $e^{i\theta}$. Moreover, it oscillates, thanks to $(-1)^k$.
2. case: $z\to\infty$. I have no clue how to proceed. I think one can make some assumptions, but I have no idea how to find the exact value of the limit.

For the record, WolframAlpha computes $0$ as the result.

Answer 1

First of all we should understand what is the meaning of the expression $z\to \infty$ when $z\in \mathbb{C}$. Note that there is no $+$ or $-$ in front of $\infty$ as one sees for limits on $\mathbb{R}$. In general the expression $z\to \infty$ means effectively $|z| \to \infty$. This $\infty$ is sometimes referred to as the point at infinity. I am assuming this is the case here. As usual one says that the limit exists if it is the same no matter the way it is approached.

Set $z=x+iy$ with $x,y\in\mathbb{R}$. $z^2 = x^2-y^2 +2i x y$. So

$$e^{-z^2} = e^{-(x^2-y^2)}e^{-2ixy}.$$

Now, if, for example, we fix $y$ and send $x\to \pm \infty$, $e^{-z^2} \to +\infty$. On the other hand fixing $y$ and sending $x\to \pm \infty$ one has $e^{-z^2} \to 0$. So the limit as $|z| \to \infty$ does not exist.

In general the behavior of $e^{-z^2}$ when $z\to \infty$ is controlled by the sign of $\mathrm{Re}(z^2)= x^2-y^2$. $x^2-y^2>0$ is the region shaded region in the following plot ($x^2-y^2<0$ is the white region). One can approach $|z|\to \infty$ in a way such that $x^2-y^2 \to +\infty$ in which case $e^{-z^2}\to 0$.

$\hspace{3cm}$

So for example one has:

• If $z \to \infty$ when $x^2-y^2 \to +\infty$, $e^{-z^2} \to 0$.
• If $z \to \infty$ when $x^2-y^2 \to -\infty$, $\left | e^{-z^2} \right | = e^{y^2-x^2} \to +\infty$.
• One the lines $y=\pm x$, $e^{-z^2} = e^{\mp 2 i y^2}$ oscillates.

But these are not the only ways one can approach the point at infinity. Indeed one can have (thanks to @SangchulLee for pointing this out) $|z| \to \infty $ on the hyperbolae $x^2-y^2 = C$ ($C>0$ corresponds to hyperboale inside the shaded region, while the $C<0$ ones are inside the white region) and on these lines $\lim_{z\to \infty} e^{-z^2} = e^{-C^2}$

Answer 2

Here is a cool theorem that might help you. If $f$ has an isolated singularity at $z_0$. $z_0$ is an essential singularity if and only if there exist sequences $\alpha_n,\beta_n$ such that both $\alpha_n,\beta_n \to z_0$ as $n\to \infty$ and $f(\alpha_n) \to 0$ and $f(\beta_n) \to \infty$.

Also $z\to\infty$ just means that $|z|\to \infty$. It makes sense when you consider the Riemann sphere.

Answer 3

You can write the complex variable in polar form. Let $z = re^{i\theta}$ where $r,\theta\in\mathbb{R}$. Then, $$ \lim_{z\to\infty}e^{-z^2}\\ =\lim_{z\to\infty}e^{-\left(re^{i\theta}\right)^2}\\ =\lim_{z\to\infty}e^{-r^2\left(e^{i\theta}\right)^2}\\ =\lim_{z\to\infty}e^{-r^2e^{2i\theta}}\\ =\lim_{z\to\infty}\left(e^{-r^2}\right)^{e^{2i\theta}} $$ $z\to\infty$ is the same as $r\to\infty$ because $\left|z\right|=\left|r\right|$ and $\left|e^{i\theta}\right|=1

$$=\lim_{r\to\infty}\left(e^{-r^2}\right)^{e^{2i\theta}}$$

We know that $\lim_{r\to\infty}e^{-r^2}=0$ in the real case, so

$$=\lim_{t\to0}t^{e^{2i\theta}}$$

This limit is also 0, hence

$$\lim_{z\to\infty}e^{-z^2}=0$$

Answer 4

We have the following sufficient (but not necessary) condition:

$$ e^{-z_n^2} \to 0 \quad\text{whenever}\quad \left|\operatorname{Re}(z_n)\right| \to \infty \text{ and } \limsup_{n\to\infty} \left| \frac{\operatorname{Im}(z_n)}{\operatorname{Re}(z_n)}\right| < 1. \tag{*} $$

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Remark. In order to see why we need extra conditions for this, consider the sequence $z_n = \sqrt{n\pi/2}(1 + i)$ and note that

$$ e^{-z_n^2} = e^{-n\pi i} = (-1)^n $$

does not converge. (Also note that $\left| \frac{\operatorname{Im}(z_n)}{\operatorname{Re}(z_n)}\right| = 1$ in this example.) More seriously, if we choose $(z_n)$ to be $z_n = \sqrt{n\pi} i$, then

$$ e^{-z_n^2} = e^{n\pi} \to \infty. $$

These two examples clearly demonstrate that $e^{-z^2}$ does not converge as $|z| \to \infty$ and we need to restrict the range of $z$ in which the limit is taken.

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Proof. Returning to proving $\text{(*)}$, write $z_n = x_n + iy_n$ with $x_n = \operatorname{Re}(z_n)$ and $y_n = \operatorname{Im}(z_n)$. Then

$$ \left| \exp(-z_n^2) \right| = \exp(\operatorname{Re}(-z_n^2)) = \exp(-(x_n^2 - y_n^2)). $$

By the assumption, there exists $r < 1$ and $N$ such that $|\frac{y_n}{x_n}| \leq r$ whenever $n \geq N$. Then $y_n^2 \leq r^2 x_n^2$, hence

$$ \left| \exp(-z_n^2) \right| \leq \exp(-(1-r^2) x_n^2) $$

for $n \geq N$. Now letting $n \to \infty$ proves the desired claim.

Answer 5

For a function of a complex variable such as $f(z)=e^{-z^2}$, $\lim_{z \to \infty} f(z)$ means the same as $\lim_{|z| \to \infty} f(z)$. In this case, no such limit exists, since $\lim_{n \to \infty} f(n)=0$ but $\lim_{n \to \infty} f(in)=+\infty.$

The limits that do exist are $\lim_{x \to +\infty} f(x)$ and $\lim_{x \to -\infty} f(x)$, which are both equal to $0$.