When does $Re(\int_Cf(z)dz) = \int_c Re(f(z)dz)$?
I genuinely have no idea where to even begin with this. I thought about letting $f$ be a function, such as $f = \frac{1}{z}$ but don't know how that would look different for each side of the equation? In other words, I guess my question is how does $Re$ change the equation?
Thank you!
That is always true. That is literally the definition of integration of a complex valued 1-form. Namely, if $\omega_1$ and $\omega_2$ are real valued 1-forms and $\omega = \omega_1 + i\omega_2$, then the definition is $$\int_{C}\omega := \int_{C}\omega_1 + i\int_{C}\omega_2.$$ This is the only definition that ensures that integration is linear.
Edit: There is some confusion on the definition of the real part $\text{Re}(\omega)$ of a complex valued 1-form. A complex valued one form $\omega$ on an open set $\Omega \subset \mathbb{R}^2$ is an object of the form $\omega = f\,dx + g\,dy$, where $f, g : \Omega \to \mathbb{C}$. In the question, we are interested in the case where $\omega = f\,dz$, where $f : \Omega \to \mathbb{C}$, and $dz = dx + i\,dy$. Write $f = u + iv$ with $u, v : \Omega \to \mathbb{R}$. Then we have $$f\,dz = (u + iv)(dx + i\,dy) = u\,dx - v\,dy + i(v\,dx + u\,dy).$$ The standard definitions of $\text{Re}$ and $\text{Im}$ of a 1-form yield $$\text{Re}(f\,dz) = u\,dx - v\,dy,$$ $$\text{Im}(f\,dz) = v\,dx + u\,dy.$$ By linearity of integration, when you integrate $f \,dz$ you get $$\int_{C} f\,dz = \int_{C}(u\,dx - v\,dy) + i\int_{C}(v\,dx + u\,dy).$$ Hence $\text{Re}(\int_{C}f\,dz) = \int_{C}\text{Re}(f\,dz)$ and $\text{Im}(\int_{C}f\,dz) = \int_{C}\text{Im}(f\,dz)$.