# $f : B_1(0) \rightarrow \mathbb{C}$ such that $f$ is injective for $\text{Re}z > 0$. What is order of pole at $0$.

Let $f : B_1(0)\setminus \{0\} \rightarrow \mathbb{C}$ be holomorphic with a pole in $0$ and assume that $f$ is injective on $B_1(0)\setminus \{0\}\cap \{z : \text{Re} z > 0\}$.

What can the order of the pole at $0$ be?

My Idea: Let $n$ be the order of the pole. Consider $g(z)=1/f(z)$. Then $g$ is also injective on the right half plane and $n$ is the order of the zero of $g$ at $0$. Now expand $g$ into a power series around $0$, $g(z) = az^n + h(z)$ with $h \in O(z^{n+1})$. Wlog $a=1$, because otherwise we can compose $g$ with a $z\mapsto \frac 1 a z$. Now for $r >0$ small enough we have $z^n < h(z)$ for $z \in \partial B_r(0)$, and for $w$ with $|w| < |z|^n - |h(z)|$ we get by Rouches theorem that $g(z)=w$ has $n$ solutions in $B_r(0)$. But unfortunately these solutions don't need to be in the right half plane. Any ideas how I could improve this argument.

I suspect the answer to be $n=1$, since then $f(z) = 1/z$ works. For higher orders I haven't yet found solutions.

• Order two $f(z)=1/z^2$ should also work. Write $f(z)=h(z)/z^n$ with $h(0)\neq0$. Argue that you can get a local $v(z)$ such that $v(z)^n=h(z)$. That should help you conclude that larger orders are not good. – Ginna Jul 24 '17 at 13:00
• yes thats true. – math635 Jul 24 '17 at 13:02
• Like this: In a small ball $B_r(0)$ $h$ is not zero. Then since the ball is simply connected there is $H$ such that $h = e^H$. Then $f(z) = (e^{H(z)/n}/z)^n$ for $z \in B_r(0)$ and therefore only $n = 1,2$ can occur? – math635 Jul 24 '17 at 13:08
• Iam not sure that we can really conclude that n=1,2 since the image of v could be contained in a sector, so that taking n'th power is invective. – math635 Jul 24 '17 at 15:15
• This problem is not as simple as it looks. – Yiorgos S. Smyrlis Jul 25 '17 at 21:52

Assume the pole is of order $n$. Then we can write $f(z)=\left(\frac{h(z)}{z}\right)^{n}$ near $z=0$ for some analytic $h$ with $h(0)\neq0$.

Let me call $\Omega$ the open upper half disc. Call $G=h(\Omega)$ and $D_a$ the image of $\Omega$ by $z\mapsto az$. Observe that $D_a$ is just another open half circle with center at the origin.

Let $a^n$ be a complex number with $|a|$ very large, such that $\frac{h(z)}{z}=a$ has solutions $z_0$ very close to $0$. Therefore $G$ and $D_a$ intersect (at $az_0$ for example).

If $n>2$, then either $D_{aw}$ or $D_{a/w}$ also intersect $G$ at $az_0$, where $w=e^{2\pi i/n}$. In fact $$D_{aw}\cup D_{a/w}\supset D_a\ \color{red}{\leftarrow\text{ This covering thing is what fails for }n\leq2}$$

Therefore, either $b=z_0/w$ (or $=z_0w$ depending on which one had the intersection) is in the open upper half unit ball, and will give us $h(b)=awb$ (or $=ab/w$). So, $f$ is not injective in the open upper half unit ball, because $f(z_0)=a^n=f(b)$.

Therefore $n\leq 2$.

The pole can be of orders $1$ or $2$ as in the examples $f(z)=\frac{1}{z}$ and $f(z)=\frac{1}{z^2}$

• Why do $G$ and $D_a$ intersect? – Yiorgos S. Smyrlis Jul 25 '17 at 21:50
• They intersect at az_0 – math635 Jul 30 '17 at 23:38

Answer. The order of the pole could be 1 or 2. For example $f(z)=z^{-1}$ and $f(z)=z^{-2}$ have the property of the OP.

Proof. Assume that the order $m$ of the pole is larger that $2$.

Construction of the reciprocal of $f$. Set $g(z)=\dfrac{1}{f(z)}$. Then $g$ is analytic in a disk $D(0,r_1)$, for some radius $0<r_1<1$ and has a unique zero at $z=0$, which is of order $m$. Also, $g$ is injective in $D(0,r_1)\cap\{\mathrm{Im}\,z>0\}$.

Construction of the $m-$th root of $g$. Since $g(z)=z^m h(z)$, where $h$ is analytic in $D(0,r_1)$ with $h(0)\ne 0$. Thus, there exists an $0<r_2<r_1$, such that $h(z)\ne 0$ in $D(0,r_2)$, which allows us to define an $m-$th root of $h$, i.e., an analytic $H$ in $D(0,r_2)$, with $H^m=h$, and if $G(z)=zH(z)$, then $G^m=g$.

The function $G$ has a single zero at $z=0$, and we can further assume, without loss of generality, that $G'(0)=1$. (Otherwise, $G$ and eventually $f$ is multiplied by a suitable constant, and the original assumption still holds.)

Inverse of $G$. Since $G'(0)\ne 0$, then $G$ possesses a holomorphic inverse $G^{-1}$ in a smaller disk $D(0,r_3)$, $0<r_3<r_2$. Let $\varrho>0$, so that $D(0,\varrho)\subset G\big(D(0,r_2)\big)$. Also $(G^{-1})'(0)=1$.

Next, take $0<\vartheta<\omega<\pi$, such that $\,m(\omega-\vartheta)=2\pi$. (For example, $\vartheta=\pi/m$ and $\omega=3\pi/m$.) Clearly, $$\alpha(t)=t\mathrm{e}^{i\vartheta},\,\beta(t)=t\mathrm{e}^{i\omega}\in D(0,\varrho)\cap\{\mathrm{Im}\,z>0\}, \quad\text{for all t\in[0,\varrho).}$$ The smooth curves $\alpha(t)$ and $\beta(t)$ intersect at $z=0$, and only there, at an angle $\omega-\vartheta$, whereas $\alpha^m(t)=\beta^m(t)$, for all $t\in[0,\varrho)$.

Claim. There exists an $\varrho_1\in (0,\varrho)$, such that $$G^{-1}\big(\alpha(t)\big),\,G^{-1}\big(\beta(t)\big)\in D(0,r_3)\cap\{\mathrm{Im}\,z>0\},$$ for all $t\in(0,\varrho_1)$.

Assuming that the claim holds, we conclude that, for all $t\in(0,\varrho_2)$, $$g\big(G^{-1}\big(\alpha(t)\big)\big)=G^m\big(G^{-1}\big(\alpha(t)\big)\big)=\Big(G\big(G^{-1}\big(\alpha(t)\big)\big)\Big)^m=\alpha^m(t)=\beta^m(t)=\cdots=g\big(G^{-1}\big(\beta(t)\big)\big),$$ and hence $g$ is NOT injective in $D(0,r_3)\cap\{\mathrm{Im}\,z>0\}\subset D(0,1)\cap\{\mathrm{Im}\,z>0\}$. Contradiction.

Proof of the Claim. Observe that $$1=G^{-1}(0)=\lim_{t\to 0}\frac{G^{-1}\big(\alpha(t)\big)}{\alpha(t)}=\lim_{t\to 0}\frac{G^{-1}\big(\alpha(t)\big)}{t\mathrm{e}^{i\vartheta}}$$ and hence $\lim_{t\to 0}\frac{G^{-1}(\alpha(t))}{t}=\mathrm{e}^{i\vartheta}=\mu+i\nu$, where $\nu>0$. Hence $$\Big|\,\mathrm{Im}\,\frac{G^{-1}(\alpha(t))}{t}-\nu\,\Big|<\frac{\nu}{2},$$ whenever $t\in (0,\tilde\varrho)$, for some $\tilde\varrho\in(0,\varrho)$. Therefore $\mathrm{Im}\,G^{-1}(\alpha(t))>0$, for all $t\in (0,\tilde\varrho)$. The case of $G^{-1}(\beta(t))$ is done in the same way. $\Box$