Show $\dfrac{g(z)-g(z_0)}{z-z_0}-f(z_0)\to 0$ as $z\to z_0$

Question

Given $f:\Omega \to \mathbb{C}$ - continuous. Define $$g(z)=\int_\rho^z f(\zeta)\,d\zeta$$ where $z\in\Omega$. Show for all $z\in\Omega$ $$\dfrac{g(z)-g(z_0)}{z-z_0}-f(z_0)\to 0$$ as $z\to z_0$

Proof: \begin{equation*} \begin{aligned} g(z)-g(z_0) & = \int_\rho^z f(\zeta)\,d\zeta - \int_\rho^{z_0} f(\zeta)\,d\zeta = \int_{z_0}^z f(\zeta)\,d\zeta \end{aligned} \end{equation*}

But from here I am unsure what to do. Any suggestions?

Answer 1

The proof is almost the same as in the real-variable case. Notice that

$$f(z_0) = \frac{1}{z - z_0} \int_{z_0}^z f(z_0) \, d\zeta$$

and so after a simplification, the difference becomes

$$\frac{1}{z - z_0} \int_{z_0}^z f(\zeta) - f(z_0) \, d\zeta.$$

Now the integrand tends to zero due to continuity, and the overall limit is zero.