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Let $S$ be the open unit square in the first quadrant of the complex plane. ($S = \{x+iy : 0 < x,y < 1\}$)

Let $f : S \rightarrow S$ be a conformap mapping.

Prove or disprove :

(1) $f$ can have no fixed point

(2) $f$ have one and only one fixed point

(3) $f$ is not an identity and $f$ have at least two fixed points

Since $S$ is simply connected, by Riemann Mapping Theorem, there exists a conformal mapping $g$ from a unit disc $D$ to $S$. Moreover, there is a result concerning fixed point of conformal mapping from $D$ onto $D$.

Precisely, if $h$ is a conformal mapping from $D$ on to $D$ with it is not an identity, then $h$ has one fixed point "inside $D$" or two fixed points on the boundary of $D$ (so no fixed points inside $D$ in this latter case)

So the latter case means no fixed points inside $D$ (exactly two on the boundary). I guess this case suggest it is possible to for $f$ to have no fixed point ?

For the case when $h$ has exactly one fixed point, it suggests that $f$ will also have $1$ fixed point ?

And finally, $h$ can have more than one fixed point only when it is an identity map from $D$ to $D$. This should means that the answer to $(3)$ is negative ?

So all the answers except (3) should be positive ? Now I try to construct examples to actually show (1) and (2) are possible.

But I have never seen such the conformal mapping before. Can anyone give some help please ?

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The problem as you stated it for $S$ is equivalent to same problem stated for $D$. And there are no automorphisms of $D$ (other than the identity) with more than one fixed point. So, the same thing holds for the automorphisms of $S$.

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