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By way of contradiction, suppose ${B}_{s/2}(f\left(c\right)$ is not contained in $f\left(U\right),$ , and let $w\in {B}_{s/2}\left(f\left(c\right)\right)$ be a complex number that is not in $f\left(U\right).$ We have that $|w-f(c\left)\right|<s/2,$ which implies that $|w-f(z\left)\right|>s/2$ for all $z\in {C}_{r}.$ Consider the function $g$ defined on the closed disk ${\overline{B}}_{r}\left(c\right)$ by $g\left(z\right)=1/(w-f(z\left)\right).$ Then $g$ is continuous on the closed disk ${\overline{B}}_{r}\left(c\right)$ and differentiable on ${B}_{r}\left(c\right).$ Moreover, $g$ is not a constant function, for if it were, $f$ would also be a constant function on ${B}_{r}\left(c\right)$ and therefore, by the Identity Theorem, constant on all of $U,$ whichg is not the case by hypothesis. Hence, by the Maximum Modulus Principle,the maximum value of $\left|g\right|$ only occurs on the boundary ${C}_{r}$ of this disk. That is, there exists a point ${z}^{\text{'}}\in {C}_{r}$ such that $\left|g\left(z\right)\right|<|g\left({z}^{\text{'}}\right)|$ for all $z\in {B}_{r}\left(c\right).$ But then
which gives the desired contradiction. Therefore, the entire disk ${B}_{s/2}\left(f\left(c\right)\right)$ belongs to $f\left(U\right),$ and hence the point $f\left(c\right)$ belongs to the interior of the set $f\left(U\right).$ Since this holds for any point $c\in U,$ it follows that $f\left(U\right)$ is open, as desired.
Now we can give the version of the Inverse Function Theorem for complex variables.
Let $S$ be a piecewise smooth geometric set, and suppose $f:S\to C$ is continuously differentiable at a point $c=a+bi,$ and assume that ${f}^{\text{'}}\left(c\right)\ne 0.$ Then:
Arguing by contradiction, suppose that $f$ is not one-to-one on any disk ${\overline{B}}_{r}\left(c\right).$ Then, for each natural number $n,$ there must exist two points ${z}_{n}={x}_{n}+i{y}_{n}$ and ${z}_{n}^{\text{'}}={x}_{n}^{\text{'}}+i{y}_{n}^{\text{'}}$ such that $|{z}_{n}-c|<1/n,$ $|{z}_{n}^{\text{'}}-c|<1/n,$ and $f\left({z}_{n}\right)=f\left({z}_{n}^{\text{'}}\right).$ If we write $f=u+iv,$ then we would have that $u({x}_{n},{y}_{n})-u({x}_{n}^{\text{'}},{y}_{n}^{\text{'}})=0$ for all $n.$ So, by part (c) of [link] , there must exist for each $n$ a point $({\widehat{x}}_{n},{\widehat{y}}_{n}),$ such that $({\widehat{x}}_{n},{\widehat{y}}_{n})$ is on the line segment joining ${z}_{n}$ and ${z}_{n}^{\text{'}},$ and for which
Similarly, applying the same kind of reasoning to $v,$ there must exist points $({\tilde{x}}_{n},{\tilde{y}}_{n})$ on the segment joining ${z}_{n}$ to ${z}_{n}^{\text{'}}$ such that
If we define vectors ${\overrightarrow{U}}_{n}$ and ${\overrightarrow{V}}_{n}$ by
and
then we have that both ${\overrightarrow{U}}_{n}$ and ${\overrightarrow{V}}_{n}$ are perpendicular to the nonzero vector $(({x}_{n}-{x}_{n}^{\text{'}}),({y}_{n}-{y}_{n}^{\text{'}})).$ Therefore, ${\overrightarrow{U}}_{n}$ and ${\overrightarrow{V}}_{n}$ are linearly dependent, whence
Now, since both $\{{\widehat{x}}_{n}+i{\widehat{y}}_{n}\}$ and $\{{\tilde{x}}_{n}+i{\tilde{y}}_{n}\}$ converge to the point $c=a+ib,$ and the partial derivatives of $u$ and $v$ are continuous at $c,$ we deduce that
Now, from [link] , this would imply that ${f}^{\text{'}}\left(c\right)=0,$ and this is a contradiction. Hence, there must exist an $r>0$ for which $f$ is one-to-one on ${\overline{B}}_{r}\left(c\right),$ and this proves part (1).
Because $f$ is one-to-one on ${B}_{r}\left(c\right),$ $f$ is obviously not a constant function. So, by the Open Mapping Theorem, the point $f\left(c\right)$ belongs to the interior of the range of $f,$ and this proves part (2).
Now write $g$ for the restriction of $f$ to the disk ${B}_{r}\left(c\right).$ Then $g$ is one-to-one. According to part (2) of [link] , we can prove that ${g}^{-1}$ is differentiable at $f\left(c\right)$ by showing that
That is, we need to show that, given an $\u03f5>0,$ there exists a $\delta >0$ such that if $0<|z-f(c\left)\right|<\delta $ then
First of all, because the function $1/w$ is continuous at the point ${f}^{\text{'}}\left(c\right),$ there exists an ${\u03f5}^{\text{'}}>0$ such that if $|w-{f}^{\text{'}}\left(c\right)|<{\u03f5}^{\text{'}},$ then
Next, because $f$ is differentiable at $c,$ there exists a ${\delta}^{\text{'}}>0$ such that if $0<|y-c|<{\delta}^{\text{'}}$ then
Now, by [link] , ${g}^{-1}$ is continuous at the point $f\left(c\right),$ and therefore there exists a $\delta >0$ such that if $|z-f(c\left)\right|<\delta $ then
So, if $|z-f(c\left)\right|<\delta ,$ then
But then,
from which it follows that
as desired.
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