If \(f\) has a simple pole at \(z_0\text{,}\) then the Laurent series is
\begin{equation*}
f(z) =\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots\text{.}
\end{equation*}
If we multiply both sides of this equation by \((z-z_0)\) and take the limit as \(z \to z_0\text{,}\) we obtain
\begin{align*}
\lim_{z \to z_0}(z-z_0) f(z) \amp = \lim_{z \to z_0}[a_{-1}+a_0(z-z_0)+a_1(z-z_0)^2+\cdots]\\
\amp =a_{-1} = \mathrm{Res}[f,z_0]\text{.}
\end{align*}
which establishes part (i). We proceed to part (iii), as part (ii) is a special case of it. Suppose that \(f\) has a pole of order \(k\) at \(z_0\text{.}\) Then \(f\) can be written as
\begin{equation*}
f(z) = \frac{a_{-k}}{(z-z_0)^k}+\frac{a_{-k+1}}{(z-z_0)^{k-1}}+\cdots +\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\cdots\text{.}
\end{equation*}
Multiplying both sides of this equation by \((z-z_0)^k\) gives
\begin{equation*}
(z-z_0)^kf(z) =a_{-k}+\cdots +a_{-1}(z-z_0)^{k-1}+a_0(z-z_0)^k+\cdots\text{.}
\end{equation*}
If we differentiate both sides \(k-1\) times we get
\begin{align*}
\frac{d^{k-1}}{dz^{k-1}}[(z-z_0)^kf(z)] \amp =(k-1) !a_{-1}+k!a_0(z-z_0)\\
\amp + {(k+1)!}\,a_1(z-z_0)^2+\cdots\text{.}
\end{align*}
and when we let \(z \to z_0\) the result is
\begin{equation*}
\lim_{z \to z_0}\frac{d^{k-1}}{dz^{k-1}}[(z-z_0)^kf(z) ] = (k-1)!a_{-1}=(k-1) !\mathrm{Res}[f,z_0]\text{,}
\end{equation*}
which establishes part (iii).
