TITLE: Can asymptotes be imaginary?
QUESTION [9 upvotes]: I'm currently learning about asymptotes and I'm trying to work out the following example:
$$f(x)=\frac{1}{x^2+1}$$
To find the vertical asymptotes, the book I'm following says that after factoring completely, you should set each factor of the denominator to $0$ and:

Every solution you get that does not make the numerator 0 will give
  you a vertical asymptote of the function.

According to that, I do:
$$
\begin{align}
x^2+1=0 \\
x^2=-1 \\
x = \pm\sqrt{-1} \\
x = \pm i 
\end{align}
$$
Now, neither $i$ nor $-i$ make the numerator $0$, so does that mean that $x=i$ and $x=-i$ are vertical asymptotes to the function $f$?  And if so, what does that really mean?
Wolfram|Alpha doesn't mention any vertical asymptotes, only the horizontal one at $y=0$.

REPLY [7 votes]: To illustrate what 3296 was getting at, consider these two plots:

These are plots of the real and imaginary parts of the function $\dfrac1{z^2+1}$, where $z=x+i y$. The poles of the function at $z=\pm i$ are easily seen in the plots. Since we're limited to seeing (a two-dimensional projection of) three dimensions, we are forced here to illustrate the poles by plotting the real and imaginary parts of the function separately.
Now, look at a "slice" of the real part:

You see at once the plot of the function you are accustomed to on the real line. The poles do not lie in the slice, and this corresponds to you seeing no vertical asymptotes in the plots of your function on the real line.
Incidentally, this function is the usual example for demonstrating the so-called "Runge phenomenon": any attempt to approximate this function with a polynomial fails due to the poles in the complex plane, even if you are only considering real values of the argument.