I have questions on the following exercise which proves Cauchy's Integral Formula for a convex region.
(Exer 4.38) This exercise gives an alternative proof of Cauchy’s Integral Formula (Theorem 4.27) that does not depend on Cauchy’s Theorem (Theorem 4.18). Suppose the region G is convex; this means that, whenever z and w are in G, the line segment between them is also in G. Suppose f is holomorphic in G, f' is continuous, and $\gamma$ is a positively oriented, simple, closed, piecewise smooth path, such that w is inside $\gamma$ and $\gamma$ $\sim_G$ 0.
(a) Consider the function $g : [0, 1] \to \mathbb C$ given by $$g(t) := \int_{\gamma} \frac{f (w +t(z−w))}{z-w} dz$$ Show that $g' = 0$. (Hint: Use Theorem A.9 (Leibniz’s rule) and then find an antiderivative for $$\frac{\partial f}{\partial t} (z + t(w − z)).)$$
(b) Prove Theorem 4.27 by evaluating g(0) and g(1).
(c) Why did we assume G is convex?
Question 1. Please verify proof for (a) and (b). zhw. already answered (c).
(a) Proof that $g'=0$:
By Leibniz' Rule A.9, $$g'(t) := \frac{d}{d t}\int_{\gamma} \frac{f(w+t(z-w))}{z-w} dz = \int_{\gamma} \frac{\partial}{\partial t} \frac{f(w+t(z-w))}{z-w} dz$$
$$= \int_{\gamma} \frac{f'(w+t(z-w))(z-w)}{z-w} dz = \int_{\gamma} f'(w+t(z-w)) dz$$
$$ = \frac{f(w+t(z-w))}{t}|_{\gamma(0)}^{\gamma(1)} = \frac{f(w+t(z-w))}{t}|_{\gamma(1)}^{\gamma(1)} = 0$$
QED that $g'=0$
(b) Proof of Cauchy's Integral Formula (Thm 4.27) on a convex region:
By Mean-Value Thm (Thm A.2), there exists $a$ in $(0,1)$ s.t. $$0 = g'(a) = g'(0+a(1)) = \frac{g(1)-g(0)}{1-0} = \frac{\int_{\gamma} \frac{f(z)}{z-w} dz-f(w)\int_{\gamma} \frac{dz}{z-w}}{1-0}$$
$$\therefore, \int_{\gamma} \frac{f(z)}{z-w} dz = f(w)\int_{\gamma} \frac{dz}{z-w} \stackrel{\forall r > 0}{=} f(w)\int_{C[w,r]} \frac{dz}{z-w} = 2 \pi if(w)$$
Therefore, we have proven Cauchy's Integral Formula (Thm 4.27) on a convex region. QED
Question 2. About the hint for (a): Why the change between $z$ and $w$? Initially, we have $f(w+t(z-w))$. Then we have $f(z+t(w-z))$.
I suspect this has something to do with the convexity where the line segment from $z$ to $w$ is given by $$\gamma_{zw}(t) := z+t(w-z)$$ while the line segment in reverse, namely the line segment from $w$ to $z$ is given by $$-\gamma_{zw}(t) := w+t(z-w) =: \gamma_{wz}(t)$$
Note that $$-\gamma_{zw}(t) := z+(1+0-t)(w-z) = z+(1-t)(w-z) = w-t(w-z) = w+t(z-w)$$ Therefore, $-\gamma_{zw}$ is indeed a reparametrisation of $\gamma_{wz}$.
Question 3. About the hint for (a): What is the relevance of computing such an antiderivative?
I computed the following respective antiderivatives of $\frac{\partial}{\partial t} f(z+t(w-z))$ and $\frac{\partial}{\partial t} f(w+t(z-w))$:
$$\frac{(w-z)f(z+t(w-z))}{1-t} + \frac{F(z+t(w-z))}{(1-t)^2} \tag{2}$$
$$\frac{(z-w)f(w+t(z-w))}{t} - \frac{F(w+t(z-w))}{t^2} \tag{3}$$
where $F$ is any antiderivative of $f$ given by the complex analogue to the Fundamental Theorem of Calculus Part I (Thm 4.15).
I don't see the point of either antiderivative.
$(2)$ Proof that the function is a required antiderivative: Observe $$\frac{d}{dz} \frac{(w-z)f(z+t(w-z))}{1-t} + \frac{d}{dz} \frac{F(z+t(w-z))}{(1-t)^2}$$
$$ = \frac{(-1)f(z+t(w-z))}{1-t} + \frac{(w-z)f'(z+t(w-z))}{1} + \frac{d}{dz} \frac{F(z+t(w-z))}{(1-t)^2}$$
$$ = \frac{(-1)f(z+t(w-z))}{1-t} + \frac{(w-z)f'(z+t(w-z))}{1} + \frac{f(z+t(w-z))}{1-t}$$
$$ = \frac{(w-z)f'(z+t(w-z))}{1} = \frac{\partial}{\partial t} f(z+t(w-z))$$
$$\therefore, \frac{(1+w-z)f(z+t(w-z))}{(1-t)^2} \ \text{is an antiderivative for} \ \frac{\partial}{\partial t} f(z+t(w-z))$$
Therefore, the function is a required antiderivative. QED
(3) similar to (2)
I am going to leave my previous answer up, to be able to refer to it.
(a) You are going through the steps I mentioned in my first answer, although there's a detail that needs attending to. We have by Leibniz that
$$\tag 1 g'(t)= \int_{\gamma} f'(w+t(z-w))\,dz.$$
Now for $t \in (0,1],$ an antiderivative for $z\to f'(w+t(z-w))$ is $f(w+t(z-w))/t.$ Any time an analytic function has an antiderivative in a region, the integral of that function over a closed contour in the region is $0.$ (This follows from the FTC.) Thus for $t \in (0,1],$ $g'(t)=0.$
What about $t=0?$ Clearly the formula for the antiderivative above doesn't work in this case. But $(1)$ is still valid and we get $g'(0)=\int_{\gamma} f'(w)\,dz.$ Here the antiderivative is $zf'(w).$ Thus $g'(0)=0.$
(b) You've made a mistake here. The MVT is for real valued functions, not complex functions, for which it can fail. But we don't need the MVT here. Just argue directly that since we now know $g$ is constant,
$$\int_{\gamma} \frac{f(z)}{z-w} dz = g(1) = g(0) = 2\pi i f(w),$$
which is what we want.
(c) This is explained in my previous answer.
Questions 2 and 3: I think your trouble here stems from typos. The book says find an antiderivative for $\partial f(z+t(w-z))/\partial t.$ As you point out, $z,w$ have been flipped. That makes no sense to you or me, so let's say they meant $\partial f(w+t(z-w))/\partial t.$ It's still wrong. I'm pretty sure that what the book is trying to say, feebly, is "Find an antiderivative for
$$\frac{\partial f(w+t(z-w))/(z-w)}{\partial t}=f'(w+t(z-w))."$$ That's exactly what we did in part (a), and it's the most natural thing in the world here. I think you can put your mind to rest on the antiderivative business now. The book messed up.