Develop $f(z)=\exp z$ in a power series in $\pi i$.
The task is fo find coefficients $a_n$ in such a way that $$ f(z)=\exp z=\sum_{n=0}^{\infty}a_n (z-\pi i)^n.~~~~~~~~~~~~~~~~~~~(1) $$ My idea is to develop $g(z):=f(z-\pi i)$ in a power series in $0$, because this power series does have the same coefficients $a_n$, because if (1), then $$ g(z)=f(z-\pi i)=\sum_{n=0}^{\infty}a_n(z-\pi i)^n.~~~~~~~~~~~~~(2) $$
Because one knows the exponential series, it is $$ g(z)=f(z-\pi i)=\exp(z-\pi i)=-\exp z=\sum_{n=0}^{\infty}\frac{1}{n!}(z-\pi i)^n\\\Leftrightarrow\exp z=-\sum_{n=0}^{\infty}\frac{1}{n!}(z-\pi i)^n $$ By comparing the coefficients in (1) and (2) it follows that the searched coefficients are $$ a_n=-\frac{1}{n!}. $$
It would be very kind of you if you gave me a short feedback whether this is okay.
With regards
math12
Your result is right, but the way you explained it is muddled. Because the series expansion$$\exp z=\sum_{n=0}^{\infty}\dfrac{1}{n!} z^n$$holds over the entire complex plane, it holds also under the replacement $z\leftarrow z-\mathrm i\pi.$ Hence$$-\exp z=\exp z\,\exp(-\mathrm i\pi)=\exp(z-\mathrm i\pi)=\sum_{n=0}^{\infty}\dfrac{1}{n!} (z-\mathrm i\pi)^n.$$So$$\exp z=-\sum_{n=0}^{\infty}\dfrac{1}{n!} (z-\mathrm i\pi)^n.$$