Consider the function $$f(z)=\frac{z}{1+z^2}$$
Clearly, $$\lim_{z\to\infty}f(z)=0$$ from any directions.
However, it still has a pole at infinity as the residue there is non zero: $$\operatorname*{Res}_{z=0}\frac1{z^2}\frac{1/z}{1+1/z^2}\ne 0$$
The function does not blow up to infinity but still has a pole there: how to reconcile? Is there an intuitive explanation for this phenomenon?
Let $w=1/z$ then $$f(z) = g(w)=\frac {w}{1+w^2}$$
$f$ has a pole at $\infty $ if and only if g has a pole at $w=0$ which is not the case.
Thus $f$ does not have a pole at $\infty$
Functions have poles; differentials have poles. The function $$\frac z{1+z^2}$$ has no pole at $\infty$ (it has a simple zero there). The differential $$\frac {z\,dz}{1+z^2}$$ has a simple pole at $\infty$. To see this, set $w=1/z$. Then $$\frac z{1+z^2}=\frac w{1+w^2}$$ and $$\frac {z\,dz}{1+z^2} =-\frac {dw}{w(1+w^2)}.$$
According to https://en.wikipedia.org/wiki/Zeros_and_poles the point at infinity is a pole of order n for $f(z)$ if for $n > 0$ $$\lim_{z\to \infty}\frac{f(z)}{z^n}$$ exists and is a non zero complex number. At this case $$\dfrac{f(z)}{z^n}=\dfrac{z^{1-n}}{1+z^2}=\dfrac{1}{z^{n-1}(1+z^2)}$$if $n\ge 1$ then the fraction tends to $0$ as $z$ tends to $\infty$ but we can say that
$\infty$ is a 1st-order zero for $\dfrac{z}{1+z^2}$
Over the complex plane, every rational function factors into linear factors. Some factors are in the numerator and some in the denominator. Consider any linear factor $\ z-c. \ $ It clearly has a simple zero at $\ z=c. \ $ We consider that it also has a simple pole at infinity to balance the simple zero. That is, the number of zeros and poles are equal. This makes sense for several reasons. If we extend this idea to all the linear factors of any rational function, then the number of zeros and poles of it are requal, up to multiplicity.
In the case of your function $\ f(z) = z/(1+z^2) = z/((z+i)(z-i)). \ $ The $\ z \ $ gives a zero at $0$ and a pole at $\infty.$ The two factors in the denominator contribute a double zero at $\infty$ along with poles at $i$ and $-i.$ Thus, in total, there are simple zeros at $0$ and $\infty$, and simple poles at $i$ and $-i.$ You can see this in another way by using partial fractions. Here $\ f(x) = \frac12(\frac1{x+i}+\frac1{x-i}). \ $ Again, there are simple poles at $i$ and $-i.$ Check that the simple zeros are at $0$ and $\infty.$
