Solution
= Solution
{c}
On the circle, the integrand in the <Cauchy derivative formula> has modulus at most $M/r^{k+1}$, and the circle has length $2\pi r$. The <ML inequality> therefore gives
$$
|f^{(k)}(z_0)|\leq\frac{k!}{2\pi}(2\pi r)\frac M{r^{k+1}}
=\boxed{\frac{k!M}{r^k}.}
$$
The maximum $M$ exists because the modulus of the continuous function $f$ attains a maximum on the compact circle. These are the <Cauchy estimate>.