\noindent 1. Over $|z| = 1,$ find the integral of $e^{2z}/z^4$

Given $f(z) = e^{2z},$ $\int_C {f(z) dz\over z^4} = {2\pi i\over 3!}
f^{(3)}(0) = {8\pi i\over 3}$

\noindent 2. Prove theorem.

$|z-z_0|$ is, on $C_R,$ exactly $R.$ Therefore, the upper bound of the
absolute value of the nth derivative of $f$ at $z_0$ becomes
$${n! \over 2*\pi*i R^{n+1}} |\int_C f(z)dz|.$$

$$|\int_C f(z) dz| < \int_C |f(z)| dz \leq 2\pi RM_R,$$
so $$|f^{(n)}(z_0)| \leq {n!M_R \over R^n}.$$

\noindent 3. If f is entire and bounded, it is a constant

$|f(z)| <= M_R$, where $M_R$ is some number.

This means the theorem from question 2 applies, and because $R$ can be
arbitrarily large, the first derivative at $z_0$ (any point in the
plane) must be bounded by $n!M_R/R \to 0$ as $R\to\infty$. This means
the function is constant.

\bye