Because
\(f^{(0)}(z) = f(z)\text{,}\) the case for
\(n=0\) reduces to
Theorem 6.5.1. We now establish the theorem for the case
\(n=1\text{.}\) We start by using the parametrization
\begin{equation*}
C:\xi =\xi (t) \quad \text{and} \quad d\,\xi = \xi \,'(t)\,dt, \quad \text{for} \quad a \le t \le b\text{.}
\end{equation*}
\begin{equation}
f(z) = \frac{1}{2\pi i}\int_C\frac{f(\xi)}{\xi -z}\,d\xi =
\frac{1}{2\pi i}\int_a^b\frac{f\big(\xi(t)\big) \xi \,'(t)\,dt}{\xi(t)-z}\text{.}\tag{6.5.6}
\end{equation}
The integrand on the right side of Equation
(6.5.6) is a function
\(g(z,t)\) of the two variables
\(z\) and
\(t\text{,}\) where
\begin{equation*}
g(z,t) = \frac{f\big(\xi(t)\big) \xi \,'(t)}{\xi(t)-z} \quad \text{and} \quad
\frac{\partial g}{dz}(z,t) = g_{z}(z,\,t) = \frac{f\big(\xi(t)\big) \xi \,'(t)}{\big(\xi (t)-z\big)^2}\text{.}
\end{equation*}
Moreover,
\(g(z,t)\) and
\(g_{z}(z,t)\) are continuous on the interior of
\(C\text{,}\) which is an open set. Applying Leibniz’s rule to Equations
(6.5.6) gives
\begin{equation*}
f\,'(z) = \frac{1}{2\pi i}\int_a^b\frac{f\big(\xi(t)\big) \xi \,'(t)\,dt}{\big(\xi(t)-z\big)^2} =
\frac{1}{2\pi i}\int_C \frac{f(\xi)\,d\xi}{(\xi -z)^2}\text{,}
\end{equation*}
and the proof for the case
\(n=1\) is complete. We can apply the same argument to the analytic function
\(f\,'\) and show that its derivative
\(f\,''\) is also represented by Equation
(6.5.5) for
\(n=2\text{.}\) The principle of mathematical induction establishes the theorem for all integers
\(n \ge 0\text{.}\)
