Can the poles of a complex function $f(z)$ be defined as the locations where $\lvert f(z) \rvert = \infty$? According to Wikipedia,

A zero of a meromorphic function $f$ is a complex number $z$ such that $f(z) = 0$. A pole of $f$ is a zero of $1/f$.

Is there a reason why a pole cannot be defined as the location where $\lvert f(z) \rvert = \infty$? I was imagining this to be the case based on this plot of the magnitude of the complex gamma function:

 A: Since $f$ is a function from $\Bbb C$ to $\mathbb C$, the range of $f$ cannot contain $\infty$. This is because $\infty\notin \mathbb C$, by definition. So, it does not make sense to say $|f(z)| =\infty$ (the modulus takes values in $[0, \infty)$ only).
Intuitively, what you say is true. In fact,

Suppose $f:\Bbb C\to\mathbb C$ has an isolated singularity at $a$. Then, $a$ is a pole of $f$ if and only if $$\lim_{z\to a} |f(z)| =\infty$$

A: In some sense the answer is no, but the reason is somewhat subtler. The issue is that a singularity of a general complex function $f$ is essentially a point where its absolute value is infinity, however, poles are only one particularly well-behaved sort of singularity, and not all singularities are poles.
For instance the function $$f(z)=e^{1/z}$$
is extremely poorly behaved at $z=0$, which you can sort of see by looking at a graph for real $z$. This is an example of an essential singularity, which is a singularity that is not a pole.
Note that this does not contradict the result mentioned in the other answer, because the limit $\lim_{z\to 0} e^{1/z}$ does not exist. Essentially, although $|f(0)|$ "should be" infinity, the limit does not actually exist because, if you pick just the right direction to approach zero (the imaginary axis), then the function actually does not blow up.
In fact, functions can have even worse singularities when they're clumped together, and these are certainly not poles. Examples of these are natural boundaries. You can see more information at the Wikipedia page.
A: I'd say that your assertion is approximately correct, since one characterization of a pole of $f$ at $z_o$ (as opposed to essential singularity) is that $\lim_{z\to z_o} |f(z)|=+\infty$. In fact, although, there are complications in talking about "the value $\infty$", it can be made to make sense for poles: the function takes values on the Riemann sphere $\mathbb C\mathbb P^1$, which is obtained by adding (in suitable fashion) a "point at infinity" to the complex plane $\mathbb C$.
The behavior of $f$ with isolated singularity that is not a pole, but is an essential singularity, is not only captured by the negation of the condition for a pole, but, further, by the Casorati-Weierstrass theorem: in every neighborhood of an essential singularity $z_o$ of $f$, $f$ comes arbitrarily near to every value in $\mathbb C$.
