Why is $$\exp(i\tau\Delta/4\pi)(xp)=x\exp(i\tau\Delta/4\pi)(p)+i\tau \exp(i\tau\Delta/4\pi)(p')/2\pi, $$ where $p$ is a polynomial. $\Delta$ denotes the Laplacian operator. I do not understand the equation. For example for $x=0$, $p$ a polynomial, it says $1=i\tau \exp(i\tau\Delta/4\pi)(p')/2\pi.$
Here $\tau$ is in the upper half plane, $p=p(x),$ $p'$ denotes the derivative of $p,$ and $$\Delta=\sum_j\frac{d^2}{dx_j^2}. $$
I will look at the 1D case. As $p$ is a polynomial, we will try exponentiating via a power series, since eventually all terms become zero. First of all, note that higher-order derivatives of a product follow the Leibniz rule: $$(fg)^{(n)}=\sum_{j=0}^n\binom{n}{j}f^{(j)}g^{(n-j)}.$$ If $f(x)=x,$ as we have here, then we get some simplifications, since second-order and higher derivatives vanish: $$(xp)^{(n)}=\sum_{j=0}^n\binom{n}{j}x^{(j)}p^{(n-j)}=xp^{(n)}+np^{(n-1)}.$$ So that's one piece. Next, we note, as Willie Wong has already mentioned in a comment, that $$e^x=\sum_{j=0}^\infty\frac{x^j}{j!}.$$ It follows that $$\exp\left(\frac{i\tau\partial^2}{4\pi}\right)=\sum_{j=0}^\infty\frac{\left(\dfrac{i\tau\partial^2}{4\pi}\right)^{\!j}}{j!}=\sum_{j=0}^\infty\frac{(i\tau)^j \partial^{2j}}{j!(4\pi)^j}, $$ but many of these terms will vanish. Let $n$ be the order of the polynomial $p$. A derivative of order $n+1$ annihilates $p$. So in the operator expansion, terms of order $n+1$ and higher will vanish. We can hence write \begin{align*} \left[\sum_{j=0}^\infty\frac{(i\tau)^j \partial^{2j}}{j!(4\pi)^j}\right]\!(xp) &=\left[\sum_{j=0}^\infty\frac{(i\tau)^j \partial^{2j}}{j!(4\pi)^j}(xp)\right]\\ &=\sum_{j=0}^\infty\frac{(i\tau)^j}{j!(4\pi)^j}\left(xp^{(2j)}+(2j)p^{(2j-1)}\right)\\ &=x\sum_{j=0}^{\lceil n/2\rceil+2}\frac{(i\tau)^j}{j!(4\pi)^j}p^{(2j)}+2\sum_{j=1}^{\lceil n/2\rceil+2}\frac{(i\tau)^j}{(j-1)!(4\pi)^j}p^{(2j-1)}\\ &=x\sum_{j=0}^{\lceil n/2\rceil+2}\frac{(i\tau)^j}{j!(4\pi)^j}p^{(2j)}+2\sum_{k=0}^{\lceil n/2\rceil+2}\frac{(i\tau)^{k+1}}{k!(4\pi)^{k+1}}p^{(2k+1)}\\ &=x\left[\sum_{j=0}^{\lceil n/2\rceil+2}\frac{(i\tau)^j \partial^{2j}}{j!(4\pi)^j}\right]\!p+\frac{i\tau}{2\pi}\left[\sum_{k=0}^{\lceil n/2\rceil+2}\frac{(i\tau)^k \partial^{2k}}{k!(4\pi)^k}\right]\!p'. \end{align*} The exact upper limit on the sums is unimportant, except that we know it is finite. Hence, rearranging sums like we have been doing is entirely justified - without worrying about convergence.
We have shown what you were trying to show.