I want to show that for holomorphic function $f\colon \mathbb{C}\to \mathbb{C}$ and any smooth map $\gamma\colon S^1\to \mathbb{C}$ $\int_{S^1}\gamma^* f(z)dz = 0$.
This fact seems very intuitive and obvious for me because the integral of holomorphic function over a closed curve equals to zero and $S^1 \subset \mathbb{R^2}$ has an obvious embedding into $\mathbb{C}$ as a set of all complex numbers such that their modulus equals to one. But what about rigorous proof?
Can it be like the following? Consider $a \in S^1$ then $\int_{S^1} \gamma^* f(z)dz = \int_{Г\colon |\gamma(a)| = 1} f(\gamma(a)) d(\gamma(a)) = 0$ because of Cauchy's integral thereom. Is it correct?
Yeah. You should interpret $f(z) \, dz$ as a complex-valued differential one form and then
$$ \gamma^{*}(f(z) \, dz) = f(\gamma(w)) d(\gamma(w)) = f(\gamma(w)) \gamma'(w) \, dw $$
leading to the expression
$$ \int_{S^1} f(\gamma(w)) \gamma'(w) \, dw = \oint_{\gamma} f(z) \, dz $$
for the integral. The right hand side vanishes because of Cauchy's theorem.