6

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 ?

PrincessEev
  • 43,815
J.Guo
  • 1,627
  • This might help: https://math.stackexchange.com/questions/398637/why-do-injective-holomorphic-functions-have-nonzero-derivative. – Martin R Feb 09 '19 at 09:33
  • @ Martin R I still don't see the point how to choose $r \le \frac{a}{c}$ , if $r$ here is fixed , then $w$ can not be sufficient small . – J.Guo Feb 09 '19 at 09:56

2 Answers2

5

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 $|z - z_0| = \epsilon$.

Martin R
  • 113,040
  • Since $F(z)=a(z-z_0)^k-w$ , $w$ is a variable of $F(z)$ , so the $\epsilon$ we choose is depending on $w$ . – J.Guo Feb 09 '19 at 09:23
  • @J.Guo: One can choose $\epsilon > 0$ first, and then $w$, see updated answer. – Martin R Feb 09 '19 at 10:32
  • I see , thank you ! – J.Guo Feb 09 '19 at 10:55
  • @MartinR, why you can choose $\varepsilon > 0$ such that $\left| \frac{G(z)}{a(z - z_0)^k} \right| < \frac{1}{2}$? I can't see why this it's true since $\left| \frac{G(z)}{a(z - z_0)^k} \right|$ is unbounded when $z \rightarrow z_0$. – George Feb 23 '19 at 13:14
  • 1
    @George: No. If $f$ takes the value $w$ with multiplicity $k$ at $z=z_0$ then $f(z)-w=a(z-z_0)^k+G(z)$ with $G(z) = O((z-z_0)^{k+1})$, so that $\frac{G(z)}{a(z - z_0)^k} \to 0$ for $z \to z_0$. – Martin R Feb 23 '19 at 13:20
  • @MartinR, is it a kind of Taylor's theorem for one complex variable? I'm search on the internet, but I just the expansion in Taylor's theorem of the function without mention about the error. If this is a kind of Taylor's theorem, can you recommend some reference for this? I studied by Conway's book, but I didn't see any mention about the error of the Taylor's series. – George Feb 23 '19 at 16:02
  • @George: Every holomorphic function can be locally developed into a power series. – Unfortunately I do not have I do not have Conway's book (or other recommendations), my university time is looong ago :) – Martin R Feb 23 '19 at 16:11
0

As stated, $G(z)=(z-z_0)^{k+1}g(z)$. Then there is some radius $r>0$ so that $$ |sg(z_0+s)|\le\frac1{6}|a|~~\text{ for all }~~ |s|\le r. $$

For some $s$ with $0<|s|\le \frac r2$ set $w=as^k$, that is, for some $|w|<|a|(r/2)^k$ set $s$ to be one of the $k$th roots of $w/a$. Then the roots of $a(z-z_0)^k-w$ are $z_0+q^js$, with $q$ the $k$th unit root, $q^k=1$. It is sufficient to consider $j=0$, for the other cases just change $s$ to $q^js$. To check that there is exactly one root of $f(z)-f(z_0)-w$ close to $z_0+s$, consider the function $$ h(z)=f(z_0+s+sz)-f(z_0)-w = as^k[(1+z)^k-1]+G(z_0+s(1+z)) $$ We will show that it has exactly one root in the disk $B(0,ϵ)$ with $ϵ=\frac2{3k}$ by showing that the first term on the right dominates the second one on the boundary circle of that disk. Note that the unit roots have a distance $2\sin(\frac\pi k)\ge 6ϵ$ for $k\ge 2$, so there is only one root $z=0$ of the first term inside the disk.

On the circle $|z|=ϵ$, we get \begin{align} |as^k[(1+z)^k-1]|&\ge |a||s|^k(kϵ-\tbinom{k}2ϵ^2-...-ϵ^k)\\ &\ge|a||s|^kkϵ\left(2-\frac1{1-\frac{k-1}2ϵ}\right)>\frac13|a||s|^k \\[1em]\hline |G(z_0+s(1+z))|&=|g(z_0+s(1+z))||s|^{k+1}|1+z|^{k+1}\\ &\le \frac16|a||s|^k(1+ϵ)^k< \frac13|a||s|^k \end{align} as $(1+\frac2{3k})^k\le e^{2/3}=1.9477..<2$. The condition for Rouché is satisfied, it follows that $h(z)$ and $as^k[(1+z)^k-1]=as^k(1+z)^k-w$ have the same number of roots in $B(0,ϵ)$, that is, exactly one.

Lutz Lehmann
  • 126,666