= Solution
Fix $R>|z|$. The <Cauchy integral formula> and the uniformly convergent geometric series on $|\zeta|=R$ give
$$
f(z)=\frac1{2\pi i}\int_{|\zeta|=R}\frac{f(\zeta)}{\zeta-z}\,d\zeta
=\sum_{n=0}^{\infty}z^n\frac1{2\pi i}\int_{|\zeta|=R}\frac{f(\zeta)}{\zeta^{n+1}}\,d\zeta.
$$
Defining
$$
c_j=\frac1{2\pi i}\int_{|\zeta|=R}\frac{f(\zeta)}{\zeta^{j+1}}\,d\zeta,
$$
the remainder after degree $N$ has modulus at most $M_R(|z|/R)^{N+1}/(1-|z|/R)$, which tends to zero uniformly on each smaller disk. Differentiating the integral formula at zero identifies $c_j=f^{(j)}(0)/j!$, independently of $R$. Since $f$ is an <entire function>, $R$ can be chosen larger than any given $|z|$, proving the global <Taylor series>.
For the requested radius $r$, the same coefficient integral gives the <Cauchy estimate>
$$
\boxed{|c_n|\le\frac{M_r}{r^n}.}
$$
To characterize <equality in the Cauchy coefficient estimate>, parametrize its circle:
$$
r^nc_n=\frac1{2\pi}\int_0^{2\pi}f(re^{i\theta})e^{-in\theta}\,d\theta.
$$
If $M_r=0$, the <identity theorem> gives $f=0$. Otherwise equality in the modulus bound requires the continuous integrand to have both constant modulus $M_r$ and constant complex argument. To see this directly, rotate its average to the positive real axis: its real part is bounded above by $M_r$, and equality of its average with $M_r$ forces that real part to equal $M_r$ everywhere. Thus $f(re^{i\theta})=c_nr^ne^{in\theta}$ on the circle. The <identity theorem> extends that equality to the plane. Conversely every such monomial attains the bound. Therefore
$$
\boxed{\text{equality holds exactly for }f(z)=cz^n,\quad c\in\mathbb C,}
$$
including the zero function and, when $n=0$, arbitrary constant functions.
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