1Complex numbers

IA Vectors and Matrices

1.2 Complex exponential function
Exponentiation was originally defined for integer powers as repeated multiplica-
tion. This is then extended to rational powers using roots. We can also extend
this to any real number since real numbers can be approximated arbitrarily
accurately by rational numbers. However, what does it mean to take an exponent
of a complex number?
To do so, we use the Taylor series definition of the exponential function:
Definition (Exponential function). The exponential function is defined as
exp(z) = e
z
= 1 + z +
z
2
2!
+
z
3
3!
+ ··· =
X
n=0
z
n
n!
.
This automatically allows taking exponents of arbitrary complex numbers.
Having defined exponentiation this way, we want to check that it satisfies the
usual properties, such as
exp
(
z
+
w
) =
exp
(
z
)
exp
(
w
). To prove this, we will
Lemma.
X
n=0
X
m=0
a
mn
=
X
r=0
r
X
m=0
a
rm,m
Proof.
X
n=0
X
m=0
a
mn
= a
00
+ a
01
+ a
02
+ ···
+ a
10
+ a
11
+ a
12
+ ···
+ a
20
+ a
21
+ a
22
+ ···
= (a
00
) + (a
10
+ a
01
) + (a
20
+ a
11
+ a
02
) + ···
=
X
r=0
r
X
m=0
a
rm,m
This is not exactly a rigorous proof, since we should not hand-wave about
infinite sums so casually. But in fact, we did not even show that the definition of
exp
(
z
) is well defined for all numbers
z
, since the sum might diverge. All these
will be done in that IA Analysis I course.
Theorem. exp(z
1
) exp(z
2
) = exp(z
1
+ z
2
)
Proof.
exp(z
1
) exp(z
2
) =
X
n=0
X
m=0
z
m
1
m!
z
n
2
n!
=
X
r=0
r
X
m=0
z
rm
1
(r m)!
z
m
2
m!
=
X
r=0
1
r!
r
X
m=0
r!
(r m)!m!
z
rm
1
z
m
2
=
X
r=0
(z
1
+ z
2
)
r
r!
Again, to define the sine and cosine functions, instead of referring to “angles”
(since it doesn’t make much sense to refer to complex “angles”), we again use a
series definition.
Definition (Sine and cosine functions). Define, for all z C,
sin z =
X
n=0
(1)
n
(2n + 1)!
z
2n+1
= z
1
3!
z
3
+
1
5!
z
5
+ ···
cos z =
X
n=0
(1)
n
(2n)!
z
2n
= 1
1
2!
z
2
+
1
4!
z
4
+ ···
One very important result is the relationship between exp, sin and cos.
Theorem. e
iz
= cos z + i sin z.
Alternatively, since sin(z) = sin z and cos(z) = cos z, we have
cos z =
e
iz
+ e
iz
2
,
sin z =
e
iz
e
iz
2i
.
Proof.
e
iz
=
X
n=0
i
n
n!
z
n
=
X
n=0
i
2n
(2n)!
z
2n
+
X
n=0
i
2n+1
(2n + 1)!
z
2n+1
=
X
n=0
(1)
n
(2n)!
z
2n
+ i
X
n=0
(1)
n
(2n + 1)!
z
2n+1
= cos z + i sin z
Thus we can write z = r(cos θ + i sin θ) = re
.