# Complex analysis primitive of a form: clarification about a proof

I have two very very basic question about some proof.

Theorem: A $C^1$ function $F$ is a primitive for $f(z) ~dz$ if and only if $F'=f$.

Proof: $F$ is a primitive of $f(z) ~dz$ $\Leftrightarrow dF = F_z ~dz + F_{\overline z } ~ d\overline z = f(z) \Leftrightarrow$ \begin{align*} F_{\overline z} &= 0 \\ F_z &= f(z) \end{align*} But (here comes the two questions) $F_z =F'$ so $F'= f(z)$.

The two questions are:

(1) How can the author guarantee that $F$ is holomorphic (has derivative $F'$)?

(2) If $F'$ exists, it is always true that $F_z = F'$. Remember that $F_z$ is defined by: $F_z = \frac12 (F_x + iF_y)$.

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What's your definition of $F'$? – Dylan Moreland Dec 3 '11 at 0:03
I have edited the question. Please make sure that this is what you intended to write. Also, what is your question in item (2)? – Srivatsan Dec 3 '11 at 0:38

You mess up with $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\overline{z}}$. The correct definition is $\frac{\partial}{\partial z}=\frac{1}{2}(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})$ and $\frac{\partial}{\partial \overline{z}}=\frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})$. (The way for me to remember is: $\frac{\partial}{\partial z}(z)=\frac{1}{2}(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})(x+iy)=1$, and $\frac{\partial}{\partial \overline{z}}=\overline{\frac{\partial}{\partial z}}$) Hence, $F_z=\frac{1}{2}(F_x-iF_y)$ and $F_{\overline{z}}=\frac{1}{2}(F_x+iF_y)$. (See here)

For question (1), we know that $F$ is holomorphic because $F_{\overline{z}}=0$. To see this, note that $F_{\overline{z}}=\frac{1}{2}(F_x+iF_y)$. So if we write $F=U+iV$, we have $F_x=U_x+iV_x$ and $F_y=U_y+iV_y$, which implies that $$F_{\overline{z}}=\frac{1}{2}(F_x+iF_y)=\frac{1}{2}(U_x+iV_x+iU_y-V_y).$$ Therefore, if $F_{\overline{z}}=0$, by the above equality $$U_x=V_y\mbox{ and }U_y=-V_x,$$ i.e. $F=U+iV$ satifised the Cauchy-Riemann equation. So $F$ is holomorphic.

For question (2), the answer is yes. If $F'$ exists, i.e. $F$ is holomorphic, then $F'$ should be equal to $F_z$.

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I interpret your question as follows: We are given a complex one-form $$\omega:=a(x,y)dx + b(x,y) dy$$ of the special form $$\omega=f(z)\ dz\ ,\quad{\rm i.e.,}\quad a=f\ , \ b = i f \qquad(1)$$ for some complex-valued function $f=u+iv$.

Such a form may or may not be the differential of a complex valued function $F$ of the variables $x$ and $y$, called a primitive (or "potential") of $\omega$. The function $F=U+iV$ is a primitive of $\omega$ iff $$F_x=a=f\ , \quad F_y=b=i f\qquad (2)$$ or $$U_x+iV_x=u+iv\ ,\qquad U_y+iV_y=-v + i u\ .$$ In particular one necessarily has $U_x=u=V_y$ and $V_x=v=-U_y$.

It follows that such an $F$ would satisfy the Cauchy-Riemann equations and so would have to be a holomorphic function of the complex variable $z=x+iy$ . In this case $f=F_x=F'(z)$, whence this is possible only if $f$ is holomorphic to begin with.

Conversely: If we have $(1)$ with a holomorphic $f$ then locally $f=F'$ for a holomorphic function $F$, and $F_x=f$, $F_y=i\, f$ as required by $(2)$.

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