Multiplicity of zeros Can you explain me how to get the multiplicity of a zero? 
In particular, I would ask you how to determine the zeros' multiplicity of
$$\cos(\frac{\pi}{2}z)$$
I suppose they are $z = 2k+1, k \in \mathbb{Z}$ and are simple zeros for the function.
 A: If $f(z) = f '(z) = 0$, $f ''(z)\ne0$ , then $z$ is a double zero.
If $f(z) = f '(z) = f ''(z) = 0$, $f '''(z)\ne0$ , then $z$ is a triple zero.
And so on.
Simple zeros are characterized with $f '(z) \ne0$.
This is the case in your example.
A: It suffices to show that the zero at $z=1$ is simple, i.e. that
$$\frac{\cos(\pi/2 z)}{z-1}$$ is a non-zero constant when $z\rightarrow 1$.
To do that, we use L'Hospital's rule to compute the limit to be $-\pi/2 \sin(\pi/2)=-\pi/2\not=0$.
For the other zeros this works the same way.
In general, an analytic function $f$ has a zero of multiplicity $k$ at $z_0$ if there exists an analytic function $g$ defined in some neighbourhood of $z_0$ with $g(z_0)\not=0$ such that
$$f(z)=(z-z_0)^k g(z)$$
close to $z_0$. In our example, the above quotient is that function $g$ and the calculation we did was to check that $g(z_0)\not=0$.
A: To find the multiplicity of a zero of $f$ calculate the logarithmic derivative $\frac{f'}{f}$ at the point.  Try it for $z^n$ at $0$!
A: The multiplicity of a zero $z$ of a function $f$ is the number $n$ such that 
$$\lim_{x\to z}\frac{f(x)}{(x-z)^n}\quad\text{is finite},$$
providing that the limit exists.
(By "finite", I mean not zero and not infinite.)
Of course it is not always defined. 
In your case, since (with $z=2k+1$)
$$\cos\left(\frac\pi2x\right)\underset{x\to z}\sim\cos\left(\frac\pi2z\right)+(x-z)
\sin\left(\frac\pi2z\right)+O\left((x-z)^2\right)$$
by Taylor expansion, and that $\sin((2k+1)\pi/2)=(-1)^k\neq0$, one can conclude that these zeros are simple.
