Residue contradiction

[Zermelo]

Hi all, I'm studying Complex Analysis and encountered a problem with calculating a residue of a function.

I'm supposed to calculate [imath]Res_{z=0} \ z\ cos^2 \frac{\pi}{z}[/imath].

I noticed that [math]0 \leq | z \cos^2 \frac{\pi}{z}| = | z|| \cos^2 \frac{\pi}{z}| \leq |z| \rightarrow 0 (z \rightarrow 0) => \lim_{z \rightarrow 0} z \cos^2 \frac{\pi}{z} = 0[/math]thus, z=0 is a removable singularity.
That means that [imath]Res_{z=0} \ z\ cos^2 \frac{\pi}{z} = 0[/imath], because the Laurent series has no singular coefficients ([imath]c_{-k} = 0,\ \forall k\in N[/imath]).

But in my class, by expanding the function into a Laurent series, we concluded that [math]c_{-1} = -\pi ^2[/math] (c-1 is the coefficient of 1/z in the Laurent series expansion), thus the residue is -pi ^2, and that's the result that Wolfram Alpha also provides.

What's wrong with my line of thought?

 

[Aligatores]

hm, I can't figure out as well(
we are missing something for sure

 

[Zermelo]

hm, I can't figure out as well(
we are missing something for sure

This is the first time a question is unanswered for so long ?
The Laurent expansion is good, there must be something wrong with the limit. Maybe [imath]\lim_{z \rightarrow 0} | f(z) | = 0[/imath] doesn't necessarily mean that [imath]\lim_{z \rightarrow 0} f(z) = 0[/imath]?
But WolframAlpha evaluates that same limit to 0! This is so crazy!
Maybe the theorem that states that a function with a removable singularity has Res = 0, has other assumptions that this function doesn't satisfy?
It also could be that this limit exisits if f is a real function, but doesn't if f is complex. But I can't see any reason why though.

UPDATE:
Cosine isn't bounded in the complex plane! Thus this limit doesn't exist! Case solved