Arguing by contradiction, suppose that ff is not one-to-one on any disk B¯r(c).B¯r(c).
Then, for each natural number n,n, there must exist two points zn=xn+iynzn=xn+iyn and zn'=xn'+iyn'zn'=xn'+iyn' such that
|zn-c|<1/n,|zn-c|<1/n,|zn'-c|<1/n,|zn'-c|<1/n, and f(zn)=f(zn').f(zn)=f(zn').
If we write f=u+iv,f=u+iv, then we would have that
u(xn,yn)-u(xn',yn')=0u(xn,yn)-u(xn',yn')=0 for all n.n.
So, by part (c) of (Reference),
there must exist for each nn a point (x^n,y^n),(x^n,y^n), such that (x^n,y^n)(x^n,y^n) is on the line segment
joining znzn and zn',zn', and for which
0
=
u
(
x
n
,
y
n
)
-
u
(
x
n
'
,
y
n
'
)
=
t
i
a
l
u
t
i
a
l
x
(
x
^
n
,
y
^
n
)
(
x
n
-
x
n
'
)
+
t
i
a
l
u
t
i
a
l
y
(
x
^
n
,
y
^
n
)
(
y
n
-
y
n
'
)
.
0
=
u
(
x
n
,
y
n
)
-
u
(
x
n
'
,
y
n
'
)
=
t
i
a
l
u
t
i
a
l
x
(
x
^
n
,
y
^
n
)
(
x
n
-
x
n
'
)
+
t
i
a
l
u
t
i
a
l
y
(
x
^
n
,
y
^
n
)
(
y
n
-
y
n
'
)
.
(2)Similarly, applying the same kind of reasoning to v,v,
there must exist points (x˜n,y˜n)(x˜n,y˜n) on the segment joining znzn to zn'zn' such that
0
=
t
i
a
l
v
t
i
a
l
x
(
x
˜
n
,
y
˜
n
)
(
x
n
-
x
n
'
)
+
t
i
a
l
v
t
i
a
l
y
(
x
˜
n
,
y
˜
n
)
(
y
n
-
y
n
'
)
.
0
=
t
i
a
l
v
t
i
a
l
x
(
x
˜
n
,
y
˜
n
)
(
x
n
-
x
n
'
)
+
t
i
a
l
v
t
i
a
l
y
(
x
˜
n
,
y
˜
n
)
(
y
n
-
y
n
'
)
.
(3)If we define vectors U→nU→n and V→nV→n by
U
→
n
=
(
t
i
a
l
u
t
i
a
l
x
(
x
^
n
,
y
^
n
)
,
t
i
a
l
u
t
i
a
l
y
(
x
^
n
,
y
^
n
)
)
U
→
n
=
(
t
i
a
l
u
t
i
a
l
x
(
x
^
n
,
y
^
n
)
,
t
i
a
l
u
t
i
a
l
y
(
x
^
n
,
y
^
n
)
)
(4)and
V
→
n
=
(
t
i
a
l
v
t
i
a
l
x
(
x
˜
n
,
y
˜
n
)
,
t
i
a
l
v
t
i
a
l
y
(
x
˜
n
,
y
˜
n
)
)
,
V
→
n
=
(
t
i
a
l
v
t
i
a
l
x
(
x
˜
n
,
y
˜
n
)
,
t
i
a
l
v
t
i
a
l
y
(
x
˜
n
,
y
˜
n
)
)
,
(5)then we have that both U→nU→n and V→nV→n are perpendicular to the nonzero vector ((xn-xn'),(yn-yn')).((xn-xn'),(yn-yn')).
Therefore, U→nU→n and V→nV→n are linearly dependent, whence
det
(
(
tial
u
tial
x
(
(
x
^
n
,
y
^
n
)
tial
u
tial
y
(
x
^
n
,
y
^
n
)
tial
v
tial
x
(
x
˜
n
,
y
˜
n
)
tial
v
tial
y
(
x
˜
n
,
y
˜
n
)
)
)
=
0
.
det
(
(
tial
u
tial
x
(
(
x
^
n
,
y
^
n
)
tial
u
tial
y
(
x
^
n
,
y
^
n
)
tial
v
tial
x
(
x
˜
n
,
y
˜
n
)
tial
v
tial
y
(
x
˜
n
,
y
˜
n
)
)
)
=
0
.
(6)Now, since both {x^n+iy^n}{x^n+iy^n} and {x˜n+iy˜n}{x˜n+iy˜n} converge to the point c=a+ib,c=a+ib,
and the partial derivatives of uu and vv are continuous at c,c, we deduce that
det
(
(
tial
u
tial
x
(
a
,
b
)
tial
u
tial
y
(
a
,
b
)
tial
v
tial
x
(
a
,
b
)
tial
v
tial
y
(
a
,
b
)
)
)
=
0
.
det
(
(
tial
u
tial
x
(
a
,
b
)
tial
u
tial
y
(
a
,
b
)
tial
v
tial
x
(
a
,
b
)
tial
v
tial
y
(
a
,
b
)
)
)
=
0
.
(7)Now, from (Reference), this would imply that f'(c)=0,f'(c)=0,
and this is a contradiction.
Hence, there must exist an r>0r>0 for which ff is one-to-one on B¯r(c),B¯r(c),
and this proves part (1).
Because ff is one-to-one on Br(c),Br(c),ff is obviously not a constant function. So, by the
Open Mapping Theorem, the point f(c)f(c) belongs to the interior of the range of f,f,
and this proves part (2).
Now write gg for the restriction of ff to the disk Br(c).Br(c).
Then gg is one-to-one.
According to part (2) of (Reference),
we can prove that g-1g-1 is differentiable at f(c)f(c) by showing that
lim
z
→
f
(
c
)
g
-
1
(
z
)
-
g
-
1
(
f
(
c
)
)
z
-
f
(
c
)
=
1
f
'
(
c
)
.
lim
z
→
f
(
c
)
g
-
1
(
z
)
-
g
-
1
(
f
(
c
)
)
z
-
f
(
c
)
=
1
f
'
(
c
)
.
(8)That is, we need to show that, given an ϵ>0,ϵ>0,
there exists a δ>0δ>0 such that if 0<|z-f(c)|<δ0<|z-f(c)|<δ then
|
g
-
1
(
z
)
-
g
-
1
(
f
(
c
)
)
z
-
f
(
c
)
-
1
f
'
(
c
)
|
<
ϵ
.
|
g
-
1
(
z
)
-
g
-
1
(
f
(
c
)
)
z
-
f
(
c
)
-
1
f
'
(
c
)
|
<
ϵ
.
(9)First of all, because the function
1/w1/w is continuous at the point f'(c),f'(c), there exists an ϵ'>0ϵ'>0 such that if
|w-f'(c)|<ϵ',|w-f'(c)|<ϵ', then
|
1
w
-
1
f
'
(
c
)
|
<
ϵ
.
|
1
w
-
1
f
'
(
c
)
|
<
ϵ
.
(10)Next, because ff is differentiable at c,c, there exists
a δ'>0δ'>0 such that if 0<|y-c|<δ'0<|y-c|<δ' then
|
f
(
y
)
-
f
(
c
)
y
-
c
-
f
'
(
c
)
|
<
ϵ
'
.
|
f
(
y
)
-
f
(
c
)
y
-
c
-
f
'
(
c
)
|
<
ϵ
'
.
(11)Now, by (Reference), g-1g-1 is continuous at the point f(c),f(c),
and therefore there exists a δ>0δ>0 such that if |z-f(c)|<δ|z-f(c)|<δ
then
|
g
-
1
(
z
)
-
g
-
1
(
f
(
c
)
|
<
δ
'
.
|
g
-
1
(
z
)
-
g
-
1
(
f
(
c
)
|
<
δ
'
.
(12)So, if |z-f(c)|<δ,|z-f(c)|<δ, then
|
g
-
1
(
z
)
-
c
|
=
|
g
-
1
(
z
)
-
g
-
1
(
f
(
c
)
)
|
<
δ
'
.
|
g
-
1
(
z
)
-
c
|
=
|
g
-
1
(
z
)
-
g
-
1
(
f
(
c
)
)
|
<
δ
'
.
(13)But then,
|
f
(
g
-
1
(
z
)
)
-
f
(
c
)
g
-
1
(
z
)
-
c
-
f
'
(
c
)
|
<
ϵ
'
,
|
f
(
g
-
1
(
z
)
)
-
f
(
c
)
g
-
1
(
z
)
-
c
-
f
'
(
c
)
|
<
ϵ
'
,
(14)from which it follows that
|
g
-
1
(
z
)
-
g
-
1
(
f
(
c
)
)
z
-
f
(
c
)
-
1
f
'
(
c
)
|
<
ϵ
,
|
g
-
1
(
z
)
-
g
-
1
(
f
(
c
)
)
z
-
f
(
c
)
-
1
f
'
(
c
)
|
<
ϵ
,
(15)as desired.
