Checking convolution property II of z transform
The term-by-term product of the sequences $(a_0, a_1, a_2)$ and
$(b_0, b_1, b_2)$ is $(c_0, c_1, c_2) = (a_0b_0, a_1b_1, a_2b_2)$. The corresponding $z$-transforms are
\begin{align}
A(z) &amp;= a_0 + a_1z^{-1} + a_2z^{-2}\\
B(z) &amp;= b_0 + b_1z^{-1} + b_2z^{-2}\\
C(z) &amp;= a_0b_0 + a_1b_1z^{-1} + a_2b_2z^{-2}
\end{align}
According to the formula in Wikipedia pointed out to you by MattL, $C(z)$ can be obtained from $A(z)$ and $B(z)$ via a contour integral in
the complex plane.  The integrand of the contour integral (which
is with respect to the complex variable $v$) is
\begin{align}
v^{-1}A(v)B\left(\frac zv\right) 
&amp;= v^{-1}\left(a_0 + a_1v^{-1} + a_2v^{-2}\right)
\left(b_0 + b_1\left(\frac zv\right)^{-1} 
+ b_2\left(\frac zv\right)^{-2}\right)\\
&amp;= \frac {a_0b_0 + a_1b_1z^{-1} + a_2b_2z^{-2}}{v}
+ \sum_{m=0}^2\sum_{n=0, n\neq m}^2 
\left(a_mb_nz^{-n}\right) v^{-m + n -1}\tag{1}
\end{align}
where in that last sum in $(1)$, the exponent of $v$ in each term is either $\geq 0$ or $\leq -2$. In other words, if we regard each of the $8$ terms in that sum as a separate function of $v$, then the function
either has no poles inside the unit circle ($-m+n-1 \geq 0$) or 
has a pole of multiplicity
greater than $1$ at the origin ($-m+n-1 \leq -2$). 
Now, write the contour integral of the right side of
$(1)$ as the sum of $9$ different contour integrals and
apply Cauchy's integral theorem to evaluate each of these $9$ contour integrals. The conclusion is that the integrals of all the terms 
in that sum in $(1)$ integrate to $0$ while the contour integral
of $\displaystyle \frac {a_0b_0 + a_1b_1z^{-1} + a_2b_2z^{-2}}{v}$ 
(which has a single pole at the origin) gives the desired
$C(z) = a_0b_0 + a_1b_1z^{-1} + a_2b_2z^{-2}$. 
Generalization of all this to sequences of arbitrary lengths is
straightforward.

If you don't know
enough complex-variable theory to figure out why it is that the contour
integral of the first term in $(1)$ is just the numerator and why the integral of the other terms is $0$, then you
will not be able to carry out the verification that you desire.