I'm reading "Signals and Systems Using Matlab" by Chaparro. In chapter 3 the Laplace transform, I'm confused by the next example.
Consider the Laplace transform
$P(s)=\frac{1}{s}(e^s-e^{-s})$
$P(s)$ seems to have a pole at $s=0$. Its zeros are obtained by letting $e^s-e^{-s}=0$, which when multiplied by $e^s$ gives
$e^s=1=e^{j2\pi k}$
for an integer $k=0,\pm1,\pm2,...$ Thus, the zeros are $s_k=j\pi k,k=0,\pm1,\pm2,...$ Now, when $k=0$, the zero at $0$ cancels the pole at zero; therefore, $P(s)$ has only zeros, an infinite number of them, $\{j\pi k,k=\pm1, \pm2, ...\}$.
I don't understand why the zeros can cancel the poles, and I haven't find any textbook in which describing the details of Zeros-Poles cancellation.
By the way, $P(s)=\frac{(s-1)(s-2)}{(s-1)(s-3)(s-5)}$ is exactly the same with $P(s)=\frac{s-2}{(s-3)(s-5)}$, having the same ROC?
This is called a removable singularity. If you have two analytic functions $f(z),g(z)$, where $f$ has a zero of order $m$ at $z_0$ and $g$ has a zero of order $n$ at $z_0$, then $f/g$ has "a zero of order $m-n$" at $z_0$. When $m-n \geq 0$, this "singularity" from the division by zero can be removed. For $m-n>0$ this is done by redefining the function to be zero at the singularity. For $m-n=0$ this is done by setting the function to be "just the right" nonzero value at the singularity (which is given by $\lim_{z \to z_0} \frac{f(z)}{g(z)}$). For $m-n<0$ this "zero" is not a zero but is instead still a pole, which is a nonremovable singularity.
What's going on in this particular example is that even though $\frac{e^0-e^{-0}}{0}$ doesn't make sense, the function $f(s)=\begin{cases} \frac{e^s-e^{-s}}{s} & s \neq 0 \\ 2 & s=0 \end{cases}$ is analytic anyway, even at zero. It is the same idea as in taking derivatives; $\frac{(1+0)^2-1^2}{0}$ makes no sense, but $g(h)=\begin{cases} \frac{(1+h)^2-1}{h} & h \neq 0 \\ 2 & h=0 \end{cases}$ is still continuous even at zero.
For more details about this, look up "complex analysis classification of isolated singularities".