prove a holomorphic function is injective

Question

Let $f: D(0,1)\to\mathbb{C}$ be holomorphic s.t $\Re (f'(z))>0\quad\forall z\in D(0,1)$. Prove that $f$ is injective.

My attempt:
Since $f'(z)=\dfrac{\partial u}{\partial x}+i\dfrac{\partial v}{\partial x}=\dfrac{\partial v}{\partial y}-i\dfrac{\partial u}{\partial y}$ (Cauchy Riemann), we have $\dfrac{\partial u}{\partial x}>0$ and $\dfrac{\partial v}{\partial y}>0$. I don't know how to continue. Could someone help me? Thanks in advance

Got it. But now why would $f'(z)=\frac {\partial u}{\partial x}+i\frac {\partial v}{\partial x}?$ — calc ll, Mar 27 '24 at 01:48
How? The derivative isn't just with respect to $x$ (or $y$). — calc ll, Mar 27 '24 at 01:51
@insomesense $f'(z_0)$ is defined to be the unique complex number such that $\lim_{z \to 0} \frac{f(z_0 + z) - f(z_0)}{z} = f'(z_0)$, where the limit is taken over complex numbers $z$ in a disc around the origin. In particular, this limit exists, so it may be evaluated along any path ending at $z = 0$, including restricting $z$ to be purely real or purely imaginary. So indeed the formulas are correct. — CJ Dowd, Mar 27 '24 at 01:59
If $f(z) = f(w)$ but $z \neq w$, then $0 = \int_0^1 \operatorname{Re}[f'(tz + (1-t)w], dt$. This cannot happen since $\operatorname{Re}[f'] > 0$ in $D(0,1)$. — kobe, Mar 27 '24 at 03:00

score 1 · Answer 1 · answered Mar 27 '24 at 08:11

Following kobe's idea, I give the proof as following: For $z\ne w$, by $N-L$ formula, we have, $$ \int_0^1 f'(tz+(1-t)w)\, dt=\frac{f(tz+(1-t)w)}{z-w}\bigg|_{0}^{1} =\frac{f(z)-f(w)}{z-w}.$$ If $f(z)=f(w)$, then $$\int_0^1 f'(tz+(1-t)w)\, dt=0,$$ so $$\Re\left(\int_0^1 f'(tz+(1-t)w)\, dt\right)=\int_0^1 \Re(f'(tz+(1-t)w))\, dt =0,$$ This cannot happen since $\Re(f'(z))>0$.

prove a holomorphic function is injective

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