Let $f$ an analytic function in a domain $D$ such that $f''(z)=0$ for all $z\in D$. Prove that there exist constants $a,b\in \Bbb C$ such that $f(z)=az+b$.
Try:
If we consider the function $g(z)=f'(z)$, we have that $g'(z)=f''(z)=0$ in $D$. And we already have a theorem that guarantees us that $g$ is constant in $D$, that is to say that $g(z)=a$ for $a\in \Bbb C$. So, $f'(z)=a$. But I don't know how to continue from here, any help. I have tried Cauchy equations.
Let $g(z)=f''(z)$. Then the analytic $g(z)\equiv 0$ by Identity theorem suffices to say $f^{(n)}=0$ for $n\ge 2$.
Use Taylor's expansion about some $\alpha\in D$ to see
$f(z)=f(\alpha)+\frac{f'(\alpha)}{1!}(z-\alpha)+0+0+....$
$\implies f(z)=[f(\alpha)-\alpha f'(\alpha)]+f'(\alpha)z=a+bz$