Prove $|\int_{\alpha} f(z) \ dz| \geq M|z-w|$

Question

I'm given a function $f: \Omega \rightarrow \mathbb{C}$. $f$ is holomorphic on $\Omega$. And we have $Re(f) \geq M > 0$. And $\alpha(t)$ connectes $z$ and $w$. Then I need to prove that $|\int_{\alpha} f(z)\ dz| \geq M|z-w|$.
I first have $|\int_{\alpha} f(z)\ dz| \geq |\int_{\alpha} Re(f(z)) \ dz|$, since $|f(z)| \geq |Re(f(z))|$ according to the definition of module of complex numbers and Pythagorean Theorem.
Then since $Re(f) \geq M$, then $|\int_{\alpha} Re(f(z)) \ dz| \geq |\int_{\alpha} M \ dz|$.
Then since it's holomorphic, then it has primitive.
Then according to the fundamental theorem of Calculus, we have $|\int_{\alpha} M \ dz| = |M(z-w)|= M|z-w|$, since $M > 0$. Thus we have $|\int_{\alpha} f(z) \ dz| > M|z-w|$
Can anyone tell if my proof is correct? Thank you!
EDIT My instructor said he mistakenly put this wrong statement on the homework.

Answer 1

The result is not true without extra assumptions, eg $\alpha$ is the segment joining $z,w$ or $\Omega$ convex.

There are many simply connected domains $\Omega$ that have points $w_0 \ne w_1$ and a holomorphic function with $\Re f \ge M>0$ st if $F$ is an antiderivative of $f$ we have $F(w_0)=F(w_1)$, so if we take any path $\alpha \subset \Omega$ connecting them, $\int_{\alpha} f(z)dz=\int_{\alpha} F'(z)dz=F(w_1)-F(w_0)=0$ so we definitely do not have $0=|\int_{\alpha} f(z)dz| \ge M|w_0-w_1|>0$

For examples of such see my answer to a previous question.