If a continuous function satisfies $|f(z)^2-1|<1$ for every $z$, then either $|f(z)-1|<1$ of $|f(z)+1|<1$ for every $z$

Question

Suppose a continuous function $f:D\rightarrow \mathbb{C}$ where $D$ is a plane domain, has the property $|f(z)^2-1|<1$ for every $z$ in $D$. Show that $|f(z)-1|<1$ of $|f(z)+1|<1$ for every $z$ in $D$.

I could not solve this but I sort of have an idea. Cleary $f(z)\neq 0$ for every $z$ in $D$. If I can define two sets $U=${$z\in D||f(z)-1|<1$} , $V=${$z\in D||f(z)+1|<1$} and show both are open and by the connectedness of $D$ I can show one of thos sets have to be empty. But I just could not achieve this. Any help will be much appreciated thanks

Answer 1

You've got the right idea.

First note that

$$U = \{z \in D \mid |f(z) - 1| < 1\} = \{z \in D \mid f(z) - 1 \in \mathbb{D}\} = \{z \in D \mid g(z) \in \mathbb{D}\} = g^{-1}(\mathbb{D})$$

where $g(z) = f(z) - 1$. As $f$ is continuous, so is $g$, and as $\mathbb{D}$ is open, $U$ is open. Likewise, you can show that $V$ is open.

The first thing you need to do is show that $U$ and $V$ are disjoint. Suppose $z \in U\cap V$, then

$$2 = |1 + 1| = |(f(z)+1) - (f(z)-1)| \leq |f(z)+1| + |f(z) - 1| < 1 + 1 = 2$$

which is a contradiction. Therefore $U\cap V = \emptyset$.

What you need to do now is show that $D = U\cup V$. One way to do this is to suppose $z \in D\setminus(U\cup V)$, so $z \in D\setminus U$ and $z \in D\setminus V$. Therefore $|f(z) - 1| \geq 1$ and $|f(z) + 1| \geq 1$ so

$$|f(z)^2 - 1| = |(f(z) - 1)(f(z) + 1)| = |f(z) - 1||f(z) + 1| \geq 1$$

which is a contradiction. Therefore $D = U\cup V$.

Therefore, the connected set $D$ is the union of two disjoint open sets, so as you've noted, one of them must be empty (and the other must be $D$).

Answer 2

Letting $\Bbb D(z_0;r)$ denote the open disk centered at $z_0$ of radius $r$, you have $U = f^{-1}(\Bbb D(1;1))$ and $V = f^{-1}(\Bbb D(-1;1))$. These sets are disjoint since $\Bbb D(1;1)$ is disjoint from $\Bbb D(-1;1)$. Since $f$ is continuous and the disks $\Bbb D(1;1)$ and $\Bbb D(-1;1)$ are open, $U$ and $V$ are open. So $\{U,V\}$ is a separation of $D$ unless $U = D$ or $V = D$. This translates to $|f(z) - 1| < 1$ for all $z\in D$ or $|f(z) + 1| < 1$ for all $z\in D$.

Answer 3

One of the $U$ and $V$ must be non empty. WLOG assume $U$ is non empty. We will show that $V$ is empty. Assume otherwise.

Take a point $z_1$ in $U$ and a point $z_2$ in $V$. Then connect those two points by a path $p$. The function $g: p \rightarrow \mathbb{C}$ defined $ z \rightarrow |f(z)-1| $ is continuous.

We have that $g(z_1)<1$ and $g(z_2)>1$ so there is $ z_0$ such that $g(z_0)=1$. From the hypothesis we get that $ | f(z_0)^2 -1| <1$ therefore $|f(z_0) +1|<1$. But $ | f(z_0) +1| = | f(z_0) -1 +2|\geq 2-| f(z_0)-1 |=1$ a contradiction.

Answer 4

Let $W=\{z| |z^2-1| < 1 \}$. Since $|z^2-1|=|z-1||z+1|$, we see that $W = B(-1,1) \cup B(1,1)$. We are given that $f(D) \subset W$.

We see that $\operatorname{re} B(-1,1) = (-2,0)$, $\operatorname{re} B(1,1) = (0,2)$, and so $\operatorname{re} W = (-2,0)\cup (0,2)$.

Let $I = (\operatorname{re} \circ f) (D) \subset \operatorname{re} W = (-2,0)\cup (0,2)$. Since $D$ is connected, we see that $I$ is connected (that is, an interval), hence we must have $I \subset (-2,0)$ or $I \subset (0,2)$, and so we must have $f(D) \subset B(-1,1)$ or $f(D) \subset B(1,1)$.