= Solution
Define a candidate <antiderivative> by integration along the straight segment from zero:
$$
g(z)=\int_{[0,z]}f(w)\,dw.
$$
For $z$ in the open unit disc and sufficiently small $h$, the triangle with vertices $0,z,z+h$ lies within the disc, since the disc is convex. By the <Cauchy integral theorem>, its boundary <contour integral> vanishes. Consequently
$$
g(z+h)-g(z)=\int_{[z,z+h]}f(w)\,dw
=h\int_0^1 f(z+th)\,dt.
$$
For $h\ne0$, subtract $f(z)$ after division by $h$. The absolute value of the remainder is at most $\sup_{0\leq t\leq1}|f(z+th)-f(z)|$, which tends to zero by continuity of the <holomorphic function> $f$ at $z$. Hence the complex <derivative> exists and \b[$g'(z)=f(z)$ for every point of the disc]. This proves that $g$ is a <holomorphic function> and supplies the requested primitive directly from the <Cauchy integral theorem>.
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