Proposition : If $f: U \to V$ is holomorphic and injective , then $f'(z) \neq 0$ for all $z \in U$ .
Proof : We argue by contradiction , and suppose that $f'(z_0) = 0$ for some $z_0 \in U$ . Then $$f(z)-f(z_0)=a(z-z_0)^k+G(z) \,\,\,\,\,\,\,\, \text{for all $z$ near $z_0$ ,}$$ with $a\neq 0 , k \ge 2$ and $G$ vanishing to order $k+1$ at $z_0$ . For sufficiently small $w$ , we write $$f(z)-f(z_0)-w=F(z)+G(z) \,\,\,\,\,\,\,\, \text{where $F(z)=a(z-z_0)^k-w$ .} $$ Since $|G(z)|\lt |F(z)|$ on a small circle centered at $z_0$ , and $F$ has at least two zeros inside that circle , Rouche's theorem implies that $f(z)-f(z_0)-w$ has at least two zeros there , a contradiction .
My question :
Why $F$ has at least two zeros inside that small circle ? We only know that $F$ has $k$ zeros in $C$ or for some large circle centered at $z_0$ . However , since $w$ is fixed , the radius $r$ of the small circle which satisfy $|G(z)|\lt |F(z)|$ can not be sufficiently large . So , how to deduce de desired conclusion by the proof given above ?
Best Answer
For $z \ne z_0$ and arbitrary $w$ we have $$ |a(z-z_0)^k-w| - |G(z)| \ge |a(z-z_0)^k| - |w| - |G(z)| \\ = |a(z-z_0)^k| \left( 1 - \left|\frac{G(z)}{a(z-z_0)^k}\right| \right) - |w| \, . $$ Now choose $\epsilon > 0$ such that $$ \left|\frac{G(z)}{a(z-z_0)^k}\right| < \frac 12 $$ for $0 < |z - z_0| \le \epsilon$. Then $$ |a(z-z_0)^k-w| - |G(z)| \ge \frac 12 |a(z-z_0)^k| - |w| $$ so that $|G(z)| < |a(z-z_0)^k-w|$ if $|w| < \frac 12 |a| \epsilon^k$ and $|z - z_0| = \epsilon$.