Type of singularity of a complex function I am trying to find the type of singularity the function
$$f(z) = \exp\left(\frac{(\cos z-1)^2}{z^4}\right).$$
has at $z = 0$.
The expression reduces to 
$$\exp\left(\frac{\sin^4(z/2)}{z^4/2}\right).$$
The function has removable singularity at $z=0$ as
$$\lim_{z\rightarrow 0}zf(z)=0.$$
Am I right?
 A: Let our function be $\exp(g(z))$, where $g(z)=\frac{4\sin^4(z/2)}{z^4}$.
Note that 
$$g(z)=\frac{1}{4}\left(\frac{\sin(z/2)}{z/2}\right)^4.$$
But 
$$\lim_{z\to 0}\frac{\sin(z/2)}{z/2}=1$$
It follows that 
$$\lim_{z\to 0} \,\exp(g(z))=\exp(1/4).$$
So although there is a singularity at $0$, it is indeed removable by defining a new function which agrees with our function everywhere (except $0$ of course) and is $\exp(1/4)$ at $0$.
Remark: The limit of $zf(z)$ is indeed $0$, but that has no real connection with the fact that $f(z)$ has a removable singularity at $0$.
A: I don't understand:
$$\cos z=\cos\left(\frac{z}{2}+\frac{z}{2}\right)=\cos^2\frac{z}{2}-\sin^2\frac{z}{2}=1-2\sin^2\frac{z}{2}\Longrightarrow (\cos z-1)^2=4\sin^4\frac{z}{2}\Longrightarrow$$
$$\Longrightarrow e^\frac{(\cos z-1)^2}{z^4}=e^{\frac{\sin^4z/2}{\left(z^2/2\right)^2}}=e^{g(z)}\,,\,\,\text{with}$$
$$g(z)=\frac{4\sin^4z/2}{z^4}=\frac{1}{4}\left(\frac{\sin z/2}{z/2}\right)^4$$
Thus, 
$$e^{g(z)}\xrightarrow [z\to 0]{} e^{1/4}\Longrightarrow \lim_{z\to 0}\,\,z\,e^{g(z)}=0$$
A: Using the power series representation:
$$\cos(z)-1 = -z^2/2 +z^4/4! - \cdots$$
$$(\cos(z)-1)^2 = z^4/4 -z^6/4! +z^8/320 - \cdots$$
$$(\cos(z)-1)^2/z^4 = 1/4 -z^2/4! +z^4/320 - \cdots$$
$$\lim_{z \to 0} ((\cos(z)-1)^2/z^4) = 1/4 $$
$$\lim_{z \to 0} f(z) = \exp(1/4) $$
Perhaps it is also needed to do statements on the convergence of the occuring series, but I think it is sufficient to say one time, that $\cos(x)$ is entire...
