6Complex power series
IA Analysis I
6.1 Exponential and trigonometric functions
Definition (Exponential function). The exponential function is
e
z
=
∞
X
n=0
z
n
n!
.
By the ratio test, this converges on all of C.
A fundamental property of this function is that
e
z+w
= e
z
e
w
.
Once we have this property, we can say that
Proposition. The derivative of e
z
is e
z
.
Proof.
e
z+h
− e
z
h
= e
z
e
h
− 1
h
= e
z
1 +
h
2!
+
h
2
3!
+ ···
But
h
2!
+
h
2
3!
+ ···
≤
|h|
2
+
|h|
2
4
+
|h|
3
8
+ ··· =
|h|/2
1 − |h|/2
→ 0.
So
e
z+h
− e
z
h
→ e
z
.
But we must still prove that e
z+w
= e
z
e
w
.
Consider two sequences (
a
n
)
,
(
b
n
). Their convolution is the sequence (
c
n
)
defined by
c
n
= a
0
b
n
+ a
1
b
n−1
+ a
2
b
n−2
+ ··· + a
n
b
0
.
The relevance of this is that if you take
N
X
n=0
a
n
z
n
!
N
X
n=0
b
n
z
n
!
and
N
X
n=0
c
n
z
n
,
and equate coefficients of z
n
, you get
c
n
= a
0
b
n
+ a
1
b
n−1
+ a
2
b
n−2
+ ··· + a
n
b
0
.
Theorem. Let
P
∞
n=0
a
n
and
P
∞
n=0
b
n
be two absolutely convergent series, and
let (
c
n
) be the convolution of the sequences (
a
n
) and (
b
n
). Then
P
∞
n=0
c
n
converges (absolutely), and
∞
X
n=0
c
n
=
∞
X
n=0
a
n
!
∞
X
n=0
b
n
!
.
Proof.
We first show that a rearrangement of
P
c
n
converges absolutely. Hence
it converges unconditionally, and we can rearrange it back to
P
c
n
.
Consider the series
(a
0
b
0
) + (a
0
b
1
+ a
1
b
1
+ a
1
b
0
) + (a
0
b
2
+ a
1
b
2
+ a
2
b
2
+ a
2
b
1
+ a
2
b
0
) + ··· (∗)
Let
S
N
=
N
X
n=0
a
n
, T
N
=
N
X
n=0
b
n
, U
N
=
N
X
n=0
|a
n
|, V
N
=
N
X
n=0
|b
n
|.
Also let
S
N
→ S, T
N
→ T, U
N
→ U, V
N
→ V
(these exist since
P
a
n
and
P
b
n
converge absolutely).
If we take the modulus of the terms of (
∗
), and consider the first (
N
+ 1)
2
terms (i.e. the first
N
+1 brackets), the sum is
U
N
V
N
. Hence the series converges
absolutely to UV . Hence (∗) converges.
The partial sum up to (
N
+ 1)
2
of the series (
∗
) itself is
S
N
T
N
, which
converges to ST. So the whole series converges to ST .
Since it converges absolutely, it converges unconditionally. Now consider a
rearrangement:
a
0
b
0
+ (a
0
b
1
+ a
1
b
0
) + (a
0
b
2
+ a
1
b
1
+ a
2
b
0
) + ···
Then this converges to
ST
as well. But the partial sum of the first 1 + 2+
···
+
N
terms is c
0
+ c
1
+ ··· + c
N
. So
N
X
n=0
c
n
→ ST =
∞
X
n=0
a
n
!
∞
X
n=0
b
n
!
.
Corollary.
e
z
e
w
= e
z+w
.
Proof. By theorem above (and definition of e
z
),
e
z
e
w
=
∞
X
n=0
1 ·
w
n
n!
+
z
1!
w
n−1
(n − 1)!
+
z
2
2!
w
n−2
(n − 2)!
+ ··· +
z
n
n!
· 1
e
z
e
w
=
∞
X
n=0
1
n!
w
n
+
n
1
zw
n−1
+
n
2
z
2
w
n−2
+ ··· +
n
n
z
n
=
∞
X
n=0
(z + w)
n
by the binomial theorem
= e
z+w
.
Note that if (
c
n
) is the convolution of (
a
n
) and (
b
n
), then the convolution
of (
a
n
z
n
) and (
b
n
z
n
) is (
c
n
z
n
). Therefore if both
P
a
n
z
n
and
P
b
n
z
n
converge
absolutely, then their product is
P
c
n
z
n
.
Note that we have now completed the proof that the derivative of e
z
is e
z
.
Now we define sin z and cos z:
Definition (Sine and cosine).
sin z =
e
iz
− e
−iz
2i
= z −
z
3
3!
+
z
5
5!
−
z
7
7!
+ ···
cos z =
e
iz
+ e
−iz
2
= 1 −
z
2
2!
+
z
4
4!
−
z
6
6!
+ ··· .
We now prove certain basic properties of
sin
and
cos
, using known properties
of e
z
.
Proposition.
d
dz
sin z =
ie
iz
+ ie
−iz
2i
= cos z
d
dz
cos z =
ie
iz
− ie
−iz
2
= −sin z
sin
2
z + cos
2
z =
e
2iz
+ 2 + e
−2iz
4
+
e
2iz
− 2 + e
−2iz
−4
= 1.
It follows that if x is real, then |cos x| and |sin x| are at most 1.
Proposition.
cos(z + w) = cos z cos w −sin z sin w
sin(z + w) = sin z cos w + cos z sin w
Proof.
cos z cos w −sin z sin w =
(e
iz
+ e
−iz
)(e
iw
+ e
−iw
)
4
+
(e
iz
− e
−iz
)(e
iw
− e
−iw
)
4
=
e
i(z+w)
+ e
−i(z+w)
2
= cos(z + w).
Differentiating both sides wrt z gives
−sin z cos w −cos z sin w = −sin(z + w).
So
sin(z + w) = sin z cos w + cos z sin w.
When
x
is real, we know that
cos x ≤
1. Also
sin
0 = 0, and
d
dx
sin x
=
cos x ≤
1. So for
x ≥
0,
sin x ≤ x
, “by the mean value theorem”. Also,
cos
0 = 1,
and
d
dx
cos x
=
−sin x
, which, for
x ≥
0, is greater than
−x
. From this, it follows
that when
x ≥
0,
cos x ≥
1
−
x
2
2
(the 1
−
x
2
2
comes from “integrating”
−x
, (or
finding a thing whose derivative is −x)).
Continuing in this way, we get that for
x ≥
0, if you take truncate the power
series for
sin x
or
cos x
, it will be
≥ sin x, cos x
if you stop at a positive term,
and ≤ if you stop at a negative term. For example,
sin x ≥ x −
x
3
3!
+
x
5
5!
−
x
7
7!
+
x
9
9!
−
x
11
11!
.
In particular,
cos 2 ≤ 1 −
2
2
2!
+
2
4
4!
= 1 − 2 +
2
3
< 0.
Since
cos
0 = 1, it follows by the intermediate value theorem that there exists
some
x ∈
(0
,
2) such that
cos x
= 0. Since
cos x ≥
1
−
x
2
2
, we can further deduce
that x > 1.
Definition (Pi). Define the smallest x such that cos x = 0 to be
π
2
.
Since
sin
2
z
+
cos
2
z
= 1, it follows that
sin
π
2
=
±
1. Since
cos x >
0 on [0
,
π
2
],
sin
π
2
≥ 0 by the mean value theorem. So sin
π
2
= 1.
Thus
Proposition.
cos
z +
π
2
= −sin z
sin
z +
π
2
= cos z
cos(z + π) = −cos z
sin(z + π) = −sin z
cos(z + 2π) = cos z
sin(z + 2π) = sin z
Proof.
cos
z +
π
2
= cos z cos
π
2
− sin z sin
π
2
= −sin z sin
π
2
= −sin z
and similarly for others.