Math 380
Solutions to hw 5
3.2.20
Let
R
=
f
x
+
iy
: 1
x
1
and
0
y
g
and
A
=
w
:
e
1
w
e
1
and
0
Arg
(
w
)
±
:
There are three things to show: (1)
z
2
R
)
e
z
2
A
, (2) every point in
A
is the image of some point
in
R
, and (3)
e
z
is 11 on
R
.
(1)
Let
z
=
x
+
iy
belong to
R
. Then
±
1
x
1
and
0
y
Now
e
z
=
e
x
e
iy
. Thus
e
1
e
x
=
j
e
z
e
1
and
0
y
=
Arg
(
e
z
)
This means that
e
z
lies in
A:
(2)
Suppose
w
lies in
A
. Then
w
=
re
with
e
1
r
e
and
0
±
Set
z
= ln
r
+
i±:
Then
z
belongs to
R
and
w
=
e
z
:
(3)
Since
R
is contained in a horizontal strip of width less than
2
, the exponential map is 11 on
R:
3.2.23
This problem is not deep, but I think it is an amusing alternative proof of the Pythagorean identity
(8) from the last assignment. Let
f
(
z
) = sin
2
z
+ cos
2
z:
(
a
)
f
is di/erentiable on
C
plane because it is formed from sums, products, and constant mulitples of
e
iz
and
e
iz
;
both of which are di/erentiable on
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 Spring '11
 Staff
 Math, Complex number, ez, sin z cos, admissible parametrization

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