1Complex differentiation
IB Complex Analysis
1.4 Logarithm and branch cuts
Recall that the exponential function
e
z
= exp(z) = 1 + z +
z
2
2!
+
z
3
3!
+ ···
has a radius of convergence of
∞
. So it is an entire function. We have the usual
standard properties, such as e
z+w
= e
z
e
w
, and also
e
x+iy
= e
x
e
iy
= e
x
(cos y + i sin y).
So given
w ∈ C
∗
=
C \ {
0
}
, there are solutions to
e
z
=
w
. In fact, this
has infinitely many solutions, differing by adding integer multiples of 2
πi
. In
particular, e
z
= 1 if and only if z is an integer multiple of 2πi.
This means
e
x
does not have a well-defined inverse. However, we do want to
talk about the logarithm. The solution would be to just fix a particular range
of
θ
allowed. For example, we can define the logarithm as the function sending
re
iθ
to
log r
+
iθ
, where we now force
−π < θ < π
. This is all well, except it
is now not defined at
−
1 (we can define it as, say,
iπ
, but we would then lose
continuity).
There is no reason why we should pick
−π < θ < π
. We could as well require
300π < θ < 302π. In general, we make the following definition:
Definition
(Branch of logarithm)
.
Let
U ⊆ C
∗
be an open subset. A branch of
the logarithm on
U
is a continuous function
λ
:
U → C
for which
e
λ(z)
=
z
for
all z ∈ U.
This is a partially defined inverse to the exponential function, only defined
on some domain
U
. These need not exist for all
U
. For example, there is no
branch of the logarithm defined on the whole C
∗
, as we will later prove.
Example. Let U = C \R
≤0
, a “slit plane”.
Then for each
z ∈ U
, we write
z
=
r
iθ
, with
−π < θ < π
. Then
λ
(
z
) =
log
(
r
)+
iθ
is a branch of the logarithm. This is the principal branch.
On
U
, there is a continuous function
arg
:
U →
(
−π, π
), which is why we can
construct a branch. This is not true on, say, the unit circle.
Fortunately, as long as we have picked a branch, most things we want to be
true about log is indeed true.
Proposition.
On
{z ∈ C
:
z 6∈ R
≤0
}
, the principal branch
log
:
U → C
is
holomorphic function. Moreover,
d
dz
log z =
1
z
.
If |z| < 1, then
log(1 + z) =
X
n≥1
(−1)
n−1
z
n
n
= z −
z
2
2
+
z
3
3
− ··· .
Proof.
The logarithm is holomorphic since it is a local inverse to a holomorphic
function. Since e
log z
= z, the chain rule tells us
d
dz
(log z) =
1
z
.
To show that
log
(1 +
z
) is indeed given by the said power series, note that
the power series does have a radius of convergence 1 by, say, the ratio test. So
by the previous result, it has derivative
1 −z + z
2
+ ··· =
1
1 + z
.
Therefore,
log
(1 +
z
) and the claimed power series have equal derivative, and
hence coincide up to a constant. Since they agree at
z
= 0, they must in fact be
equal.
Having defined the logarithm, we can now define general power functions.
Let
α ∈ C
and
log
:
U → C
be a branch of the logarithm. Then we can define
z
α
= e
α log z
on U. This is again only defined when log is.
In a more general setting, we can view
log
as an instance of a multi-valued
function on
C
∗
. At each point, the function
log
can take many possible values,
and every time we use
log
, we have to pick one of those values (in a continuous
way).
In general, we say that a point
p ∈ C
is a branch point of a multivalued
function if the function cannot be given a continuous single-valued definition in
a (punctured) neighbourhood
B
(
p, ε
)
\ {p}
of
p
for any
ε >
0. For example, 0 is
a branch point of log.
Example. Consider the function
f(z) =
p
z(z − 1).
This has two branch points,
z
= 0 and
z
= 1, since we cannot define a square
root consistently near 0, as it is defined via the logarithm.
Note we can define a continuous branch of f on either
10
or we can just kill a finite slit by
10
Why is the second case possible? Note that
f(z) = e
1
2
(log(z)+log(z−1))
.
If we move around a path encircling the finite slit, the argument of each of
log
(
z
)
and
log
(
z −
1) will jump by 2
πi
, and the total change in the exponent is 2
πi
. So
the expression for f(z) becomes uniquely defined.
While these two ways of cutting slits look rather different, if we consider this
to be on the Riemann sphere, then these two cuts now look similar. It’s just
that one passes through the point ∞, and the other doesn’t.
The introduction of these slits is practical and helpful for many of our
problems. However, theoretically, this is not the best way to think about multi-
valued functions. A better treatment will be provided in the IID Riemann
Surfaces course.