\[\frac{{\mathrm{d}}^{n}}{{\mathrm{d}z}^{n}}\left(e^{z^{2}}\operatorname{erfc}z%
\right)=(-1)^{n}2^{n}n!e^{z^{2}}\mathop{\mathrm{i}^{n}\mathrm{erfc}}\left(z%
\right),\]