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Choose a star centre $a$ of the open star-shaped set $D$ and define the candidate antiderivative by a straight-segment contour integral:For each fixed $z\in D$ and all sufficiently small $h$, the filled triangle with vertices $a,z,z+h$ lies in $D$. Indeed, the compact segment $[a,z]$ has a neighbourhood of some positive radius contained in the open set $D$, and every point of this triangle lies within $|h|$ of that segment. The zero triangle integral consequently givesBy continuity, the last integral tends to $f(z)$ as $h\to 0$. HenceIn particular, this proves the existence of the antiderivative without incorrectly assuming the star-shaped set is convex.

On a general domain, the conclusion can fail. Take $D=\mathbb{C}\setminus \{0\}$ and $f(z)=1/z$. Every filled triangle contained in $D$ has zero boundary integral by Cauchy integral theorem, but the unit circle has integral $2\pi i$. An antiderivative would make every closed contour integral zero. This is the period obstruction to a holomorphic antiderivative: the local conclusion of Morera's theorem does not remove the global obstruction.

The hypothesis makes $\Omega$ star-shaped with centre $z_0$. For small $h$ with the triangle having vertices $z_0,w,w+h$ contained in $\Omega$, the vanishing triangular integral givesParametrizing the final segment,by continuity. Thus $g$ is holomorphic andThis is the triangle-integral construction behind Morera's theorem.

Uniform convergence on compact subsets makes $f$ continuous. For every triangle whose closure lies in $U$, uniform convergence on its boundary permits passage through the contour integral:Morera's theorem implies that $f$ is holomorphic. If a compact set $K⋐U$ is surrounded by a finite union of contours at positive distance from $K$, the Cauchy integral formula for derivatives givesThe uniform bound on the surrounding compact set proves that $f_n^{\prime }\to f^{\prime }$ uniformly on $K$.

If $f$ had distinct zeros $c$ and $d$, choose disjoint small closed discs around them whose boundary contains no zero of $f$. Uniform convergence and Rouché's theorem imply that, for large $n$, $f_n$ has a zero in each disc, contradicting uniqueness of $c_n$. Thus $f$ has at most one zero.

For an example on the unit disc, takeThen $f_n\to f(z)=z-1$, which has no zero in the open disc. In general, Hurwitz's theorem shows that $f$ is zero-free exactly when the unique zeros escape every compact subset of $U$:Indeed, an interior accumulation point of $c_n$ is a zero of $f$, while a zero of $f$ forces the unique $c_n$ into each of its sufficiently small neighbourhoods.

A function $f:U\to \mathbb{C}$ is holomorphic when it is complex differentiable at every point of the open connected set $U$. Morera's theorem states that a continuous function on a domain is holomorphic if its integral around every triangle whose interior lies in the domain is zero.

The integrand $e^{tz}/(1+t^2)$ is continuous jointly in $(t,z)$ on $[0,1]\times \mathbb{C}$, so the displayed integral defines a continuous function. For any triangle $T$, Fubini's theorem and the Cauchy integral theorem giveMorera's theorem therefore proves that $f$ is entire.

The function $1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ but has no antiderivative there, because an antiderivative would integrate to zero around every closed curve whereasThis is the standard period obstruction to a holomorphic antiderivative.