Can we see this integral as the line integral of a 1-form

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In Stein and Shakarchi's complex analysis, the following definition is given on pg. 21

Let $z:[a, b]\to \mathbf C$ be a parameterization of smooth curve $\gamma$ in $\mathbf C$ and $f$ be a continuous function of $\gamma$. We define the integral of $f$ along $\gamma$ by $$\int_\gamma f(z)dz=\int_a^b f(z(t))z'(t) dt$$

I also know that the line integral of a $1$-form $\omega$ on $\mathbf C$ along $z$ is defined as $$\int_z \omega = \int_{[a, b]}z^*\omega$$ where $z^*\omega$ denotes the pullback of $\omega$ on $[a, b]$.

Is there a way to see the first definition as a special case of the second definition?

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Yes, I think so: let $\omega$ be the (complex-valued) $1$-form $f(z)dz$. In this case, the line integral of $\omega$ along a curve $\gamma:[a,b]\to \mathbb{C}$ agrees with the line integral of the function $f$ along $\gamma$ (by the relevant definitions). You can compute the pullback $\gamma^{\ast}\omega$ as $f(\gamma(t))\gamma'(t)dt$ on $[a,b]$, where $dt$ denotes the standard $1$-form on $[a,b]$, by the chain rule (and the definition of "pullback of forms").

Hope that helps!