Final answer:
The Maclaurin series expansion for f(z) = e^z about z0 = πi is given by f(z) = f(z0) + f'(z0)(z - z0) + f''(z0)/(2!)(z - z0)^2 + f'''(z0)/(3!)(z - z0)^3 + ..., where f'(z) denotes the derivative of f(z) with respect to z, f''(z) denotes the second derivative, and so on. The radius of convergence for the Maclaurin series expansion is infinite, since e^z is an entire function.
Step-by-step explanation:
The Maclaurin series expansion for f(z) = e^z about z0 = πi is given by:
f(z) = f(z0) + f'(z0)(z - z0) + rac{f''(z0)}{2!}(z - z0)^2 + rac{f'''(z0)}{3!}(z - z0)^3 + ...
where f'(z) denotes the derivative of f(z) with respect to z, f''(z) denotes the second derivative, and so on.
The radius of convergence for the Maclaurin series expansion is the distance from the center of the expansion, z0, to the nearest singularity of f(z). In this case, since e^z is an entire function (i.e., it is analytic everywhere in the complex plane), the radius of convergence is infinite.