若$f(z)$與$\phi(z)$在點$z_0$處都連續，則$f(z)+\phi(z)$,$f(z)\cdot \phi(z)$以及$\frac{f(z)}{\phi(z)}$都是連續的.
證明:$f(z)$在$z_0$處連續，意味着對於任意給定的模大於0的複數$\varepsilon_1$,都存在相應的模大於0的複數$\delta_1$,使得$\forall |z-z_0|<\delta_1$,都有
\begin{equation}
\label{eq:6.29}
|f(z)-f(z_0)|<|\varepsilon_1|
\end{equation}
$\phi(z)$在$z_0$處連續，意味着對於任意給定的模大於0的複數$\varepsilon_2$,都存在相應的模大於0的複數$\delta_2$,使得$\forall |z-z_0|<\delta_2$,都有
\begin{equation}
\label{eq:6.33}
|\phi(z)-\phi(z_0)|<|\varepsilon_2|
\end{equation}
因此
\begin{equation}
\label{eq:6.56}
|[f(z)+\phi(z)]-[f(z_0)+\phi(z_0)]|\leq |f(z)-f(z_0)|+|\phi(z)-\phi(z_0)|<|\varepsilon_1|+|\varepsilon_2|
\end{equation}
因此$f(z)+\phi(z)$在$z_0$連續.
\begin{align*}
|f(z)\phi(z)-f(z_0)\phi(z_0)|&=|f(z)\phi(z)-f(z)\phi(z_0)+f(z)\phi(z_0)-f(z_0)\phi(z_0)|\\&=|f(z)(\phi(z)-\phi(z_0))+\phi(z_0)(f(z)-f(z_0))|\\&\leq |f(z)||\phi(z)-\phi(z_0)|+|\phi(z_0)||f(z)-f(z_0)|\\&<|\varepsilon_1||f(z)|+|\varepsilon_2||\phi(z_0)|
\end{align*}
由於$f(z)$在$z_0$連續，因此當包含點$z_0$的鄰域足夠小時，$|f(z)|$不超過某個正實數.綜上可見，$f(z)\phi(z)$在$z_0$處連續.
下面證明當$\phi(z_0)\neq 0$時,
\begin{equation}
\label{eq:7.22}
\frac{1}{\phi(z)}
\end{equation}在$z_0$處是連續的.
\begin{equation}
\label{eq:7.29}
|\frac{1}{\phi(z)}-\frac{1}{\phi(z_0)}|=\frac{|\phi(z)-\phi(z_0)|}{|\phi(z)\phi(z_0)|}
\end{equation}
當包含$z_0$的鄰域足夠小時，$\phi(z)$不爲0(因爲$\phi(z_0)$不爲0，且$\phi(z)$在$z_0$連續)，因此$|\phi(z)\phi(z_0)|$足夠接近一個正實數，而$|\phi(z)-\phi(z_0)|$足夠小，因此$\frac{1}{\phi(z)}$在$z_0$處連續.