Question: Let the function $f$ be analytic in the entire complex plane, real valued on the real axis, and of positive imaginary part in the upper half-plane. Prove that $f'(x) > 0$ for $x \in R$.
This is an old Berkeley qual problem, so I actually have the answer, which is as follows:
Let $f(x + iy) = u(x,y) + iv(x,y)$. Thus, we have: $$\frac{\partial f}{\partial x}(x,0) = \frac{\partial u}{\partial x}(x,0) + i\frac{\partial v}{\partial x}(x,0) = \frac{\partial u}{\partial x}(x,0),$$ since $v = 0$ on the real axis. Now, by the Cauchy-Riemann equations, $$\frac{\partial f}{\partial x}(x,0) = \frac{\partial v}{\partial y}(x,0)\ge 0,$$ since $v = 0$ on the real axis, but $v \ge 0$ on the upper half-plane. Therefore, all we need to do is show that $f'(x)$ does not vanish on the real axis.
If we can show that $f'(0) \neq 0$, then we will be done. We proceed by contradiction: if $f'(0) = 0$, because $f$ is holomorphic at $0$, then we can write: $$f(z) = cz^k(1+O(z)), $$ where $k \ge 2$, and $c \neq 0$ is real. If $z$ is small, the arguments of $1 + O(z)$ is between $-\pi/4$ and $\pi/4$, say, and on any half circle in the upper half plane, centered at $0$, $cz^k$ assumes all possible arguments. On a sufficiently small such half-circle, then, the product assumes arguments between $\pi$ and $2\pi$, and thus $\Im f(z)$ is not greater than $0$ for all $\Im z > 0$. This is a contradiction, and thus we are done.
I have quite a few questions on this proof that I do not understand. Here they are:
1) Why is it true that if $f'(0)\neq 0$, then $f'(x) \neq 0$ on $R$? I am assuming that it has something to do with analytic continuation, but I am not sure.
2) Why can we write $f(z) = cz^k (1+ O(z))$? I trie writing $f$ as a power series centered at $0$ to show this, but it didn't work out right.
3) Why are the arguments of $1 + O(z)$ between $-\pi/4$ and $\pi/4$? By arguments, I am assuming it means thus usual $\theta$ in the $z = re^{i\theta}$ expression, but I'm actually not sure, since I have no idea why that would mean $\theta$ is between $-\pi/4$ and $\pi/4$.
Thanks.
To answer 1, there is nothing in this argument that is particular to zero. The only thing that will change is that for any $a\in R$, you will be expanding $f$ in a neighborhood of $a$ and hence the terms in your power series will be $z-a$. When $z$ is close to $a$, the higher powers in your expansion will be dominated by $z-a$. You can proceed analogously...
To answer 2, you write $$f(z) = cz^k + a_{k+1}z^{k+1} + \ldots = cz^k(1 + \frac{a_{k+1}}{c}z + \ldots) = cz^{k+1}(1 + O(z))$$ where you get $O(z)$ because you are looking in a neighborhood of zero, and the $z$ term will dominate all higher powers of $z$ when $|z| < 1$.