Exchanging the limit of a contour integral

Question

If $f(z)$ is analytic/holomorphic on a curve $C$ and $f_n$ is uniformly convergent on $C$ does

$$\lim_{n\to\infty} \oint_{C}^{ } f_ndz = \oint_{C}^{ } \lim_{n\to\infty} f_ndz$$

I've seen to theorems relating to the real version of this problem, namely:

and

Would the same requirements be sufficient for the contour integral in the complex plane?

I've seen that this property is proven using Morera's Theorem however this assumes that $f$ is holomorphic within the region enclosed by $C$.

Answer 1

Let $c:[a,b] \to \mathbb C$ a parametrization of $C$. Then, with $g_n:= (f \circ c)c'$:

$\oint_{C}^{ } f_ndz = \int_a^b g_n(t) dt$

Now split in real - and imaginary part to use the above results for the real case.

Answer 2

If $|f_n(z) - f(z)| \leqslant \varepsilon$ for all $z \in C$, then

$$\left| \int \limits_C f_n(z) \, \mathrm{d} z - \int \limits_C f(z) \, \mathrm{d} z \right| \leqslant \int \limits_C |f_n(z)-f(z)| \, \mathrm{d} z \leqslant |C| \cdot \varepsilon$$

thus

$$\lim_{n \to \infty} \int \limits_C f_n(z) \, \mathrm{d} z = \int \limits_C f(z) \, \mathrm{d} z.$$