MATH3403 Functions of a Complex Variable
The Exponential Function
4 October, 2016
Lecture Notes 8
x1 Denition using real coordinates
Let be a real number. De Moivre's formula gives (cos + i sin )n = cos(n) +
i sin(n )
for any natural number n. If we introduce the notation ei := cos + i sin ,
then, de Moivre's formula is given by (ei )n = eni .
For z = x + iy 2 C, we dene
f (z )
= ex eiy = ex(cos y + i sin y) = ex cos y + iex sin y = u + iv
where u(x; y) = ex cos y; v(x; y) = ex sin y.
We see that u and v are functions on R admitting continuous partial derivatives of
2
all orders. Furthermore,
8
>
@u
>
>
<
= e cos y;
@u
>
@v
>
>
:
= ex sin y;
@v
@x
@x
x
@y
@y
2
=
ex sin y
= ex cos y :
;
In other words,
8
>
@u
>
>
<
@x
=
@v
@y
>
@u
>
>
:
Thus, the function f (z ) =
;
@v
=
:
@y
@x
u(x; y )+ iv (x; y ) satises the system of Cauchy-Riemann
equations and dene a complex-analytic function and we write f (z ) = ez = exp(z ).
We verify that for z = x + iy; z 0 = x0 + iy0,
e z ez
0
=
ex eiy
ex
0
eiy =
0
ex ex
0
eiy eiy
0
= ex
+x
0
eiy eiy
Furthermore, by the addition formula for sine and cosine
eiy eiy
0
= (cos y + i sin y) (cos y0 + i sin y0)
= (cos y cos y0 sin y sin y0) + i(sin y cos y0 + cos y sin y0)
= cos(y + y0) + i sin(y + y0)
= ei y
( +y 0 )
:
3
0
We have thus proven that
= ex x e i y
ez e z
0
+
0
( +y 0 )
=ex
( +x0 )+i(y +y 0 )
= ez
+z 0
:
The latter formula is a fundamental property of the exponential function.
Remark.
ez +2ki
= ez for any k 2 N and z 2 C. Therefore, ez is a periodic entire
function (function analytic on C) with period 2i.
Remark. For all z 2 C, ez e z = e0 = 1. Thus ez 6= 0 for all z 2 C.
4
x2 Denition using power series
For the time being we ignore the denition of the exponential function in x1 and dene
ez
dierently using power series. We will show afterwards that the two denitions agree
with each other.
Consider the formal power series
1
P
n=0
zn
where 0 ! = 1. From the Ratio Test the
n!
series converges absolutely for any complex number z .
In what follows recall that convergent power series f (z ) =
1
P
n=0
cn(z
analytic functions and that they can be dierentiated term by term, i.e.,
f 0 (z )
=
1
X
ncn(z
a)n
1
=
n=1
1
X
+1
n=0
5
(n + 1)cn (z
a)n:
a)n
dene
Write e =
z
1
P
zn
n=0
n
!
. Unless otherwise specied ez will carry this meaning in x2.
For z = x real, ex carries the usual meaning.
For z = iy purely imaginary we have
1
1
1
n
m
X
X
X
(
iy
)
(
iy
)
(iy) m
iy
e =
=
+
n!
(2
m) !
(2m + 1) !
n
m
m
From i m = (i )m = ( 1)m we conclude that
1
1
m m
X
X
(
1)
y
( 1)my m
iy
e =
+i
= cos y + i sin y :
(2m) !
(2m + 1) !
m
m
2
=0
2
2
=0
+1
=0
2
2
=0
2
+1
=0
The entire analytic function ez dened as above can be characterized as follows.
6
Proposition 1
Let
R >
0 and
: D(0; R) ! C be a complex-analytic function dened by a
f
power series convergent on
D (0; R).
(i) f 0(z ) f (z ) on
and (ii)
Proof.
D (0; R)
Write f (z ) =
f 0(z )
=
1
1
P
n=0
X
n=1
cn z n .
ncnz n
1
Then,
f (0)
f (z )
ez on D(0; R) if and only if
= 1.
Then
1
X
=
(m + 1)cm
m=0
Thus f 0(z ) f (z ) if and only if
1
P
n=0
cn z
n
=
+1
1
P
n=0
zm
=
1
X
(n + 1)cn
+1
z n:
n=0
(n + 1)cn
+1
zn
for all z 2 D(0; R).
From the uniqueness of coecients of power series expansions it follows that f 0(z ) f (z )
if and only if cn = (n + 1)cn for all natural numbers n.
+1
7
This happens if and only if
cn
1 cn
1
1 cn
c
cn =
=
=
= =
:
n+1
n+1
n
n+1
n
n 1
(n + 1) !
In other words, f 0(z ) f (z ) if and only if
+1
1
f (z )
2
=c
1
X
0
= c ez
!
n
= f (0), we conclude that f (z ) = ez if and only if furthermore f (0) = 1.
0
=0
Since c
zn
0
We proceed to prove the identity e e = e
z w
n
z +w
0
for the denition e =
For this purpose we will need the following lemma on series.
8
z
1
P
zn
n=0 n
!
.
Lemma 1 Let
1
P
m=0
m
be an absolutely convergent series of complex numbers. Let
: N ! N be a bijective mapping. Then
1
X
(n)
=
n=0
1
X
m :
m=0
Proof (Optional).
For a natural number M write sM =
We have to prove that Nlim
t =
!1 N
Since
M (")
1
P
m=0
1
P
M
P
m=0
m=0
m
and tN =
N
P
n=0
(n)
for partial sums.
m .
jmj is convergent, given any " > 0 there exists a natural number M =
such that
1
P
m=M +1
jmj < ".
Since : N ! N is a bijection, for each natural number M there exists a natural
number N such that f(1); ; (N )g f1; 2; ; M g.
0
0
9
In other words, it follows that for any N N , we have
0
N
X
(n)
n=0
M
X
m=0
m
1
X
jmj < "; i.e. jtN
j
sM < " :
m=M +1
Taking limits as M ! 1 we conclude that
lim
t = lim s =
!1 N M !1 M
N
1
X
m;
m=0
as asserted.
Lemma 1 applies equally well to double series. Using Lemma 1 we deduce
Proposition 2
Write
e
and
w
z
z
=
1
P
zn
n=0
n
we have
!
on the complex plane C. Then, for any two complex number
= ez
ez e w
+w
.
10
Write ez ew =
1
P
zn
1
P
wm
.
!
m
1
P
j
j
j
j
The double series
is convergent with sum ejzj ejwj.
n !m !
n
m
1 P
1 z nwm
P
in any order we like. In
By Lemma 1 we can sum the double series
n
!
m
!
n
m
particular we can sum rst over all pairs of indices (n; m) such that n + m = p and
Proof.
=0
n=0 n !
1 znw
P
m=0 m
=0
=0
=0
then sum over p.
Thus,
e
z
e =
w
1
X
0
@
p=0
1
z k w`
X
k +`=p
k !`
!
A
=
1
X
p=0
1
p
0
!
@
X
k +`=p
By the Binomial Theorem
X
k +`=p
p
!
k !`
!
z k w`
11
= (z + w)p :
p
!
k !`
!
1
z k w`A
It follows that
e
z
e =
w
1
X
p=0
1
p
!
(z + w)p = ez
+w
;
as asserted.
We can now deduce the two denitions of the exponential functions agree.
Proposition 3
For z = x+iy;
Then
f (z )
Proof.
x; y
2 R, dene f (z) = e cos y+ie sin y and dene g(z) =
x
x
= g(z ).
1
P
zn
n=0
n
!
.
By Proposition 2 we have g(z ) = g(x + iy) = g(x)g(iy).
For x 2 R we have g(x) = ex, while for iy 2 iR we have g(iy) = cos y + i sin y, as
explained.
It follows that g(z ) = ex(cos y + i sin y) = f (z ), as asserted.
Remark.
ez +2ki
= ez for any k 2 N and z 2 C.
12
x3 The exponential and logarithm functions as inverse functions
We dene Log : C fx 2 R : x 0g ! C, the principal branch of (natural)
j j + iArg(z), where logjzj = lnjzj and
Arg(z ) is the argument (polar angle) of z such that < Arg(z ) < ; z 2= fx 2 R :
x 0g. We have
logarithm, by the formula
Log (z )
=
log z
Proposition 4
Let
C be the domain dened by = fz :
the domain dened by 0
function maps
< Im(z ) < g and 0 C be
= C fx 2 R : x 0g ! C. Then the exponential
bijectively onto 0. Furthermore, the inverse function is also
analytic, given by
Log
: 0 ! . In other words,
(i)
Log (ez )
=z
for all z 2 (ii)
eLog w
=w
for all w 2 0 :
13
Proof.
Let z 2 . Then, z = x + iy;
We have ez = ex
For
ez
+iy
x; y
2 R; with
< y < .
= ex eiy = ex(cos y + i sin y).
< y < ;
sin y = 0 if and only if y = 0, in which case cos 0 = 1 and
= ex > 0.
Thus, ez is never (real and) nonpositive, i.e., ez 2 0, so that exp : ! 0.
Consider now Log : 0 ! C. For w = u + iv 2 0 we have by denition Log w =
j j + iArg(w) with
log w
<
Arg(w) < . In other words, Log : 0 ! .
We proceed to prove that exp : ! 0 and Log : 0 ! are inverse functions of
each other in the sense (i) Log(ez ) = z for all z 2 , and (ii) eLog w = w for all w 2 0.
(This will imply in particular that both mappings exp : ! 0 and Log : 0 ! are bijective.)
14
We have ez = ex eiy ; jez j = ex; Arg(ez ) = y for all z 2 , so that
Log (ez )
= logjez j + iArg(ez ) = x + iy = z :
On the other hand for w = u + iv in : eiArg w =
( )
= elogjwj
eLog w
+
w
, so that
jwj
w
iArg w
= elogjwj eiArg w = jwj = w ;
jw j
( )
( )
as asserted.
Exercise.
d
Log
dz
Exercise.
Log z1
Log z2
z
= z.
1
+ Log z = Log z z if
2
1 2
< Argz1
+ Argz , and Log z +
2
1
6= Log z z otherwise.
1 2
Denition. A continuous function
f
on a domain not containing the origin is
called a branch of the logarithm on if for all z 2 , we have ef z = z . We
( )
shall denote a branch of the logarithm on by log.
15
Example. For any
n
2 Z, Log z + 2ni is a branch of the logarithm on 0 =
C fx 2 R : x 0g.
Denition. Given a branch of the logarithm log on a domain 0
C f0g and
a complex number c. Dene the complex exponential z c by z c = ec log z for any
2 . The principal branch of zc is dened by zc = ec Log z .
p
Example. z can be dened, in any domain where log z is dened, by
z
0
pz = exp 1 log z
2
p
and for any non-zero z , there are only two possible values for z as
1
1
[log z + 2ki]
exp
log z = exp
2
2
if k is even.
Example. The set of possible values for ii ei log i is f e3=2; e
16
=2
;e
5=2
;
g: