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Let $U\subseteq \mathbb{C}$ be open, let $L$ be a straight line, and let $f:U\to \mathbb{C}$ be continuous on $U$ and holomorphic on $U\setminus L$. Then $f$ is holomorphic on all of $U$. Near a point of $L$, work in a disk whose closure lies in $U$. Split a small filled triangle along $L$ and triangulate the two resulting convex polygons. Translate each triangle slightly into its corresponding open half-plane, keeping it in $U$. Its contour integral is zero by Cauchy's integral theorem. On each segment the integral is the endpoint difference times the integral of $f$ along the unit-interval parametrization; uniform continuity on a containing compact disk therefore makes the contour integral continuous as the vertices move back. Each untranslated piece consequently has zero integral. Internal oriented edges cancel when the pieces are added, so every small triangle has zero integral. Morera's theorem proves the claim locally along the line. The continuity across $L$ is essential; holomorphicity on the two sides alone is insufficient.

Morera's theorem states that if $f$ is continuous on an open set $U$ and its integral around every triangle whose interior and boundary lie in $U$ is zero, then $f$ is holomorphic on $U$.

For a proof, choose a disk $B\subseteq U$ and a base point $z_0\in B$. Define $F(z)={\int }_{[z_0,z]}f(\zeta )d\zeta$ along the straight segment. Convexity keeps every triangle with vertices $z_0,z,z+h$ inside $B$. Its zero boundary integral impliesBy continuity, the last integral tends to $f(z)$ as $h\to 0$, so $F^{\prime }(z)=f(z)$. Thus $F$ is holomorphic on $B$. The allowed result that the derivative of a holomorphic function is holomorphic now gives that $f=F^{\prime }$ is holomorphic there. Such disks cover $U$, proving

For each triangle compactly contained in $U$, part (i) permits the interchange of the series and its boundary integral. Cauchy's integral theorem makes every term zero. The uniform limit is continuous, so Morera's theorem yields that $f$ is holomorphic on $U$.

For the final application, each function $(z-k{)}^{-2}$ is holomorphic away from its pole. Fix a positive integer $n$. For $z\in U_n$ and $k>2n$,The numerical series ${\sum }_{k>2n}4/k^2$ converges, so the Weierstrass M-test gives uniform convergence of the tail on $U_n$. Adding the finitely many initial terms preserves uniform convergence. Those initial terms need not be bounded near their excluded poles; only the tail needs the uniform bound. By the first conclusion, the full sum is holomorphic on each $U_n$. Their union is $\mathbb{C}\setminus \mathbb{N}$, hence

Morera's theorem states that a continuous function on an open subset of $\mathbb{C}$ is holomorphic if its integral around every triangular contour, with triangle and interior contained in that open set, is zero.

To prove this, work in any open disk $D$ whose closure lies in $U$, and fix $z_0\in D$. Definewhere the segment lies in $D$ by convexity. The zero integral on the triangle with vertices $z_0,z,z+h$ givesContinuity at $z$ implies that the last integral tends to $f(z)$ as $h\to 0$: its difference from $f(z)$ is bounded by ${\sup }_{0\le t\le 1}|f(z+th)-f(z)|$. Therefore $F^{\prime }=f$ on $D$. The function $F$ is holomorphic, and the allowed theorem that a holomorphic derivative is holomorphic implies that $f=F^{\prime }$ is holomorphic on $D$.

Every point of $U$ lies in such a disk, proving the result throughout $U$:The proof is local and does not require $U$ to be simply connected.

First take $z\in U_{-}$ and put $w=\bar{z}\in U_{+}$. Reflection symmetry of the function givesHence $f$ is holomorphic on $U_{-}$, with $f^{\prime }(z)=\overline{f^{\prime }(\bar{z})}$ there.

It remains to cross the real axis. Fix a real point $p\in U$ and a small disk about $p$ with closure in $U$. Consider a triangle in a still smaller disk. Intersect its filled interior with the closed upper and lower half-planes and triangulate the resulting convex polygons. This gives finitely many triangles lying on one side of the real axis, possibly with edges on that axis. Degenerate pieces contribute zero.

For an upper piece, translate its three vertices by $i\epsilon$ with $\epsilon >0$. For sufficiently small $\epsilon$ the translated triangle lies entirely in $U_{+}$ and remains in $U$, so its boundary integral is zero by Cauchy's theorem. Part (d), or its segment-integral proof, shows that the integral converges to that around the untranslated piece as $\epsilon \downarrow 0$. Thus the original piece has zero integral. Translate lower pieces by $-i\epsilon$ to obtain the same conclusion in $U_{-}$.

When all the oriented triangle integrals are added, their internal edges cancel, including the segments on the real axis. The original triangle therefore has zero boundary integral. Continuity of $f$ on $U$ now allows Morera's theorem in the disk, proving holomorphicity near $p$. Together with the two half-plane arguments this givesThis is a Morera gluing lemma across a straight line; the original continuity hypothesis is essential to the limiting step.

Morera's theorem states that a continuous complex-valued function on an open set is holomorphic if its integral around every triangle whose closed interior lies in the set is zero. It is enough to impose this condition locally on disks.

Fix a disk $B$ contained in the open set and a base point $z_0\in B$. Define $F(z)$ as the integral of $f$ along the straight segment from $z_0$ to $z$. The zero triangle integral gives, for sufficiently small $h$,Continuity makes the quotient by $h$ tend to $f(z)$, so $F$ is holomorphic and $F^{\prime }=f$. To justify that its derivative is holomorphic, apply the Cauchy integral formula for $F$ on a smaller circle:The right side may be differentiated again inside the circle because its denominator stays uniformly away from zero on the contour. Thus $F^{\prime }=f$ is holomorphic on $B$. Such disks cover the open set, proving the theorem.

The uniform limit theorem first gives continuity of $f$ on $D$. For any triangle $\Gamma$ whose closed interior is in $D$, the Cauchy integral theorem gives ${\int }_{\Gamma }{f}_n(z)dz=0$. Moreover,Hence ${\int }_{\Gamma }f=0$, and Morera's theorem proves that $f$ is analytic on $D$. In fact, the proof only needs locally uniform convergence on compact subsets, since each triangle is compact; the stated uniform convergence on the entire domain is stronger.

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.

Let the common domain be an open set $U$. The uniform limit theorem makes the limit $f$ continuous. For every triangle whose closed interior lies in $U$, Cauchy's integral theorem gives ${\int }_{\partial T}{f}_n=0$. Part (ii) passes this identity to the limit, giving ${\int }_{\partial T}f=0$. Morera's theorem now proves that $f$ is holomorphic.

The same reasoning works for locally uniform convergence, since each triangle boundary is compact. No passage of derivatives through a limit has been assumed.

Morera's theorem says that a continuous complex-valued function $f$ on an open set $U\subseteq \mathbb{C}$ is holomorphic iffor every triangle $T$ whose closed interior is contained in $U$. Triangle boundaries are traversed with positive orientation. The continuity hypothesis matters: the zero-integral condition is combined with continuity, not used in isolation.

Morera's theorem states that a continuous function on a domain is holomorphic if its contour integral around every triangle whose filled-in interior lies in the domain is zero.

The uniform limit $f$ is a continuous function, by the uniform limit theorem. For any such triangle $T$, the estimation lemma givesEach integral of $f_n$ vanishes by the Cauchy integral theorem. Hence the integral of $f$ vanishes too, and Morera's theorem proves $f$ is holomorphic on $D$. This is the basic locally uniform convergence of holomorphic functions argument; the given uniform convergence on all of $D$ is more than needed.