 Solution

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