If $f$ is $C^2$, does $g(z_1):=\int_{0}^{z_1} \frac{\partial f(z)}{\partial z} dz + f(0)$ imply that $dg=\frac{\partial f}{\partial z} dz$?

Question

If $f$ is a $C^2$ function on an open set $A \subset \mathbb{C}$ containing $z=0$ and we define $$g(z_1):=\int_{0}^{z_1}F(z) dz + f(0),$$ where $z_1 \in A$ and $F(z):= \frac{\partial f}{\partial z}(z)$ is holomorphic, does this imply that

$$dg=\frac{\partial f}{\partial z} dz?$$

I saw this equivalence on a proof but I don’t know how to rigorously deduce it as I haven’t worked yet with total differentials in complex analysis. If that’s of any use, I know that we can prove that $g’(z)=f’(z)$.

What is $\int_0^z$ ?. In the complex plane we integrate on curves or contours — marwalix, Mar 30 '23 at 15:33
@marwalix I tried to fix the question. I meant the path integral from $0$ to $z_1$, where both points are in $A$. — dahemar, Mar 30 '23 at 15:39
I'm guessing it's $g(z_1)$ then? As for the integral, I'm guessing it's the integral on the line $[0,z_1]$, for example along $\Gamma : t \mapsto tz_1$. — Bruno B, Mar 30 '23 at 15:39
@BrunoB No, $g$ is being defined as the primitive of $f$ in this context. — dahemar, Mar 30 '23 at 15:41
Yes but $z$ is the integration variable, nothing outside of the integral can be called $z$. — Bruno B, Mar 30 '23 at 15:42
@BrunoB I’m trying to see what you mean, but it’s defined like that in this answer for instance: https://math.stackexchange.com/a/108134/931269 — dahemar, Mar 30 '23 at 15:51
@BrunoB As for the integral, yes, it’s what you described. — dahemar, Mar 30 '23 at 15:51
I think they just forgot to change the name of the variable (happens to everyone). See how it's "$\gamma$ path from $z_0$ to $z$". — Bruno B, Mar 30 '23 at 16:09

score 2 · Answer 1 · answered Mar 30 '23 at 16:14

Use the decomposition of the total complex derivative $d=\partial + \bar{\partial}$, i.e., $$dh(z) = \frac{\partial h}{\partial z}dz + \frac{\partial h}{\partial \bar{z}}dz $$ for every $h$ differentiable (in the sense that both partial derivative with respect to $x$ and $y$ exists where $z=x+iy$). Then remember that $$\frac{\partial h}{\partial \bar{z}}=0$$ for every holomorphic function $h$. Finally use Stokes' Theorem: $$\int_{\partial \Omega} w = \int_{\Omega}dw$$ with $w=f(z)$ and $\Omega$ the image of the curve (in $A$) that joins $0$ and $z_1$.

I made a mistake on the question, $f$ is actually just $C^2$, not holomorphic. $F$ and $g$ are holomorphic though — dahemar, Mar 30 '23 at 19:49

If $f$ is $C^2$, does $g(z_1):=\int_{0}^{z_1} \frac{\partial f(z)}{\partial z} dz + f(0)$ imply that $dg=\frac{\partial f}{\partial z} dz$?

1 Answers1