 Solution

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.