Fundamental theorem of calculus for complex analysis, proof

I've been trying to verify/fill in the details of my book's proof of the complex FTOC, but have gotten stuck. Here are the statement and proof: If a continuous function $f$ has a primitive $F$ in $\Omega$, and $\gamma$ is a curve in $\Omega$ that begins at $w_{1}$ and ends at $w_{2}$, then $\int_{\gamma}f(z)dz = F(w_{2}) - F(w_{1})$.

Here, a primitive for $f$ on $\Omega$ is a function $F$ that is holomorphic on $\Omega$ and such that $F'(z)=f(z)$ for all $z\in \Omega$, where that derivative is meant to be the complex derivative. Also, $\gamma$ being smooth means that any of its parametrizations $z:[a,b] \rightarrow \mathbb{C}$ is continuously differentiable.

Proof: Suppose we have such a parametrization, and $z(a)=w_{1},z(b)=w_{2}$. then using the chain rule and the fundamental theorem of calculus, \begin{align*} \int_{\gamma}f(z)\,dz &= \int_{a}^{b}f(z(t))z'(t)\,dt \\ &= \int_{a}^{b}F'(z(t))z'(t)\,dt \\ &= \int_{a}^{b}\frac{d}{dt}F(z(t))\,dt \\ &= F(z(b)) - F(z(a)) \end{align*}

I'm having trouble with the 3rd and 4th equalities in the above string. First, $F'$ means the complex derivative. We had a theorem earlier that said for any function $g:\mathbb{C} \rightarrow \mathbb{C}$ holomorphic at some $z_{0}$, $g'(z_{0})=\frac{\partial g}{\partial z}(z_{0})$, where we defined the operator $\frac{\partial}{\partial z} = \frac12\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right)$. Let $m,n$ be the component functions of $z$, i.e. $m,n:[a,b]\rightarrow \mathbb{R}$ such that $z(t)=m(t)+i n(t)$ for all $t\in [a,b]$. Similarly, let $U,V:\Omega \rightarrow \mathbb{C}$ denote the component functions of $F$.

Then \begin{align*} F'(z(t))z'(t) &= \frac{\partial F}{\partial z}(z(t)) \left[\frac{d}{dt}m(t) + i\frac{d}{dt}n(t)\right] \\ &= \frac{1}{2}\left[\frac{\partial F}{\partial x}(z(t)) - i\frac{\partial F}{\partial y}(z(t))\right] \left[\frac{d}{dt}m(t) + i\frac{d}{dt}n(t)\right] \\ &= \frac{1}{2}\left(\left[\frac{\partial U}{\partial x} + \frac{\partial U}{\partial y}\right] +i\left[\frac{\partial V}{\partial y} + \frac{\partial V}{\partial x}\right]\right) \left[\frac{d}{dt}m(t) + i\frac{d}{dt}n(t)\right] \end{align*} On the other hand, I'm not sure how to compute $\frac{d}{dt}F(z(t))$ using the chain rule. I'd be able to do it if I considered $F$ as a function from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$. I've tried guessing that the notational convention is \begin{align*} \frac{d}{dt}F(z(t)) &= \frac{d}{dt}U(z(t)) + i \frac{d}{dt} V(z(t)) \end{align*} where $U(z(t)),V(z(t))$ are to be computed using the multivariable(real) chain rule, but that gave me something different from what I got for $F'(z(t))z'(t)$ above.

For the fourth equality, where the FTOC is supposed to be used, I'm not sure how to parse it. The FTOC that I know is for functions $:\mathbb{R} \rightarrow \mathbb{R}$. Any hints, insight, or more detail are very much appreciated.

Solution 1:

For functions $f,g:\Bbb C\to\Bbb C$, the chain rule applies

Let $f:U\to V$ be complex differentiable at $z_0$, and let $g:V\to\mathbb C$ be complex differentiable at $w_0=f(z_0)$. Then $g\circ f$ is complex differentiable at $z_0$ and $(g\circ f)'(z_0) = g'(w_0)f'(z_0)$.

We can also replace $f$ by a continuously differentiable curve $\gamma$ in $V$, and we still get $(g\circ\gamma)'(x_0) = g'(\gamma(x_0))\gamma'(x_0)$

Now assume $f:\Omega\to\Bbb C$ is continuous and has a primitive $F$, and that $\gamma$ is a piecewise continuously differentiable curve in $\Omega$. The integral of $f$ along a curve is defined as $$ \int_\gamma f\,dz = \int_a^b f(\gamma(t))\gamma'(t)\,dt $$ If $x_0<\dots<x_n$ denote the points such that $\gamma$ is continuously differentiable on each $[x_{i-1},x_i]$, then the right side of this equation is $$ \sum_{i=1}^n \int_{x_{i-1}}^{x_i} f(\gamma(t))\gamma'(t)\,dt $$ We noticed that this is equal to $$ \sum_{i=1}^n \int_{x_{i-1}}^{x_i} (F\circ\gamma)'(t)\,dt $$ which, by definition, is equal to $$ \sum_{i=1}^n \int_{x_{i-1}}^{x_i} \mathrm{Re}(F\circ\gamma)'(t)\,dt + i \int_{x_{i-1}}^{x_i} \mathrm{Im}(F\circ\gamma)'(t)\,dt $$ Since maps from the reals to Euclidean space are differentiated coordinatewise, we get $$ \sum_{i=1}^n \int_{x_{i-1}}^{x_i} (\mathrm{Re}F\circ\gamma)'(t)\,dt + i \int_{x_{i-1}}^{x_i} (\mathrm{Im}F\circ\gamma)'(t)\,dt $$ By the fundamental theorem of calculus, this is equal to $$ \sum_{i=1}^n \mathrm{Re}F(\gamma(x_i))-\mathrm{Re}F(\gamma(x_{i-1})) + i\mathrm{Im}F(\gamma(x_i)) - i\mathrm{Im}F(\gamma(x_{i-1})) $$ which can be simplified to $$ F(\gamma(b)) - F(\gamma(a)) $$