I know that \begin{equation} \partial_x f\,dx+\partial_yf\,dy=\partial_zf\,dz+\partial_{\bar z}f\,d\bar z, \end{equation} that we can also write as \begin{equation} df=\partial f\,+\bar\partial f. \end{equation} How can I prove that \begin{equation} df=2\Re\{\partial f\} \end{equation} in case $f$ is real valued?
Thank you for your help.
Simply write out the definitions of $\partial f$, $f_z$, and $dz$ $$\partial f = f_{z} dz = \frac{1}{2} ( f_x - i f_y ) (dx + i dy) = \frac{1}{2}( f_x dx + f_y dy + i (-f_y dx + f_x dy))$$ (do note that since $f$ is real valued, then $f_x$ and $f_y$ are real valued)