Can I get feedback on my proof from Ahlfors ?
If $g(w)$ and $f(z)$ are analytic functions, show that $g(f(z))$ is also analytic
First step is to show that if $g'(w)$ and $f'(z)$ exist, then the derivative of $g(f(z))$ is $g'(f(z))\cdot f'(z)$.
Claim
$f'(z)$ exists iff there is some $A \in \mathbb{C}$ such that $f(z+h) - f(z) = Ah + o(h)$ where $\lim_{h \to 0} \frac{o(h)}{h} = 0$
Allow $f'(z) = \lim_{h \to 0} \frac{f(z+h) - f(z)}{h}$. Then we see that $0 = \lim_{h\to 0} \frac{f(z+h) - f(z) - hf'(z)}{h}$. Thus, set $o(h)$ is the numerator and we are done.
Assume the existence of such an $A$. Then we have $o(h) = f(z+h) - f(z) - hA$, and $\lim_{h\to 0} \frac{f(z+h) -f(z)}{h} = A = f'(z)$.
Now to prove the chain rule.
Allow $k = f(z+h) - f(z)$ and $w = f(z)$
So $g(f(z+h)) - g(f(z)) = g(w + k) - g(w) = kg'(w) + o(k)$
Thus $g(f(z+h)) - g(f(z)) = (f(z+h) - f(z))g'(f(z)) + o(k)$
Since $\lim_{h \to 0} k = \lim_{h \to 0} f(z+h) -f(z) = 0$, then $\lim_{h \to 0} \frac{o(k)}{h} = 0$
$$\lim_{h\to 0} \frac{g(f(z+h)) - g(f(z))}{h} - g'(f(z))\frac{f(z+h)-f(z)}{h} = 0$$
Therefore we have proved when $h(z) = g(f(z))$ that $h'(z) = g'(f(z))\cdot f'(z)$.
Now, to show that $h(z) = g(f(z))$ is analytic, we must show it satisfies the Cauchy-Riemann equations.
Allow $g(w) = g(x,y) = u(x,y) + iv(x,y)$ and $f(z) = f(x,y) = r(x,y) + is(x,y)$. Then $h(z) = g(f(x,y)) = u(r(x,y), s(x,y)) + iv(r(x,y), s(x,y))$
From there we see that $$\frac{\partial h}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$$ and $$\frac{\partial h}{\partial y} = -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}$$
Where $$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial u}{\partial s}\frac{\partial s}{\partial x} \\ \frac{\partial u}{\partial y} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial u}{\partial s}\frac{\partial s}{\partial y} \\ \frac{\partial v}{\partial x} = \frac{\partial v}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial v}{\partial s}\frac{\partial s}{\partial x} \\ \frac{\partial v}{\partial y} = \frac{\partial v}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial v}{\partial s}\frac{\partial s}{\partial y}$$
Using the fact that both $g(w)$ and $f(z)$ satisfy Cauchy-Riemann, we can immediately do substitution to see that $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$
Thus $h(z)$ is analytic.
I know this is long, but I'd love any feedback on correctness or if I'm missing any steps. Thanks