1Analytic functions

IB Complex Methods 1.2 Complex differentiation
Recall the definition of differentiation for a real function f(x):
f
0
(x) = lim
δx0
f(x + δx) f(x)
δx
.
It is implicit that the limit must be the same whichever direction we approach
from. For example, consider
|x|
at
x
= 0. If we approach from the right, i.e.
δx
0
+
, then the limit is +1, whereas from the left, i.e.
δx
0
, the limit is
1. Because these limits are different, we say that
|x|
is not differentiable at
the origin.
This is obvious and we already know that, but for complex differentiation,
this issue is much more important, since there are many more directions. We
now extend the definition of differentiation to complex number:
Definition
(Complex differentiable function)
.
A complex differentiable function
f : C C is differentiable at z if
f
0
(z) = lim
δz0
f(z + δz) f(z)
δz
exists (and is therefore independent of the direction of approach but now
there are infinitely many possible directions).
This is the same definition as that for a real function. Often, we are not
interested in functions that are differentiable at a point, since this might allow
some rather exotic functions we do not want to consider. Instead, we want the
function to be differentiable near the point.
Definition
(Analytic function)
.
We say
f
is analytic at a point
z
if there
exists a neighbourhood of
z
throughout which
f
0
exists. The terms regular and
holomorphic are also used.
Definition
(Entire function)
.
A complex function is entire if it is analytic
throughout C.
The property of analyticity is in fact a surprisingly strong one! For example,
two consequences are:
(i)
If a function is analytic, then it is differentiable infinitely many times. This
is very very false for real functions. There are real functions differentiable
N
times, but no more (e.g. by taking a non-differentiable function and
integrating it N times).
(ii) A bounded entire function must be a constant.
There are many more interesting properties, but these are sufficient to show us
that complex differentiation is very different from real differentiation.
The Cauchy-Riemann equations
We already know well how to differentiate real functions. Can we use this to
determine whether certain complex functions are differentiable? For example is
the function
f
(
x
+
iy
) =
cos x
+
i sin y
differentiable? In general, given a complex
function
f(z) = u(x, y) + iv(x, y),
where
z
=
x
+
iy
are
u, v
are real functions, is there an easy criterion to determine
whether f is differentiable?
We suppose that
f
is differentiable at
z
. We may take
δz
in any direction
we like. First, we take it to be real, with δz = δx. Then
f
0
(z) = lim
δx0
f(z + δx) f (z)
δx
= lim
δx0
u(x + δx, y) + iv(x + δx, y) (u(x, y) + iv(x, y))
δx
=
u
x
+ i
v
x
.
What this says is something entirely obvious since we are allowed to take the
limit in any direction, we can take it in the
x
direction, and we get the corre-
sponding partial derivative. This is a completely uninteresting point. Instead,
let’s do the really fascinating thing of taking the limit in the y direction!
Let δz = y. Then we can compute
f
0
(z) = lim
δy0
f(z + y) f(z)
y
= lim
δy0
u(x, y + δy) + iv(x, y + δy) (u(x, y) + iv(x, y))
y
=
v
y
i
u
y
.
By the definition of differentiability, the two results for
f
0
(
z
) must agree! So we
must have
u
x
+ i
v
x
=
v
y
i
u
y
.
Taking the real and imaginary components, we get
Proposition
(Cauchy-Riemann equations)
.
If
f
=
u
+
iv
is differentiable, then
u
x
=
v
y
,
u
y
=
v
x
.
Is the converse true? If these equations hold, does it follow that
f
is differ-
entiable? This is not always true. This holds only if
u
and
v
themselves are
differentiable, which is a stronger condition that the partial derivatives exist, as
you may have learnt from IB Analysis II. In particular, this holds if the partial
derivatives u
x
, u
y
, v
x
, v
y
are continuous (which implies differentiability). So
Proposition.
Given a complex function
f
=
u
+
iv
, if
u
and
v
are real differen-
tiable at a point z and
u
x
=
v
y
,
u
y
=
v
x
,
then f is differentiable at z.
We will not prove this proofs are for IB Complex Analysis.
Examples of analytic functions
Example.
(i) f
(
z
) =
z
is entire, i.e. differentiable everywhere. Here
u
=
x, v
=
y
. Then
the Cauchy-Riemann equations are satisfied everywhere, since
u
x
=
v
y
= 1,
u
y
=
v
x
= 0,
and these are clearly continuous. Alternatively, we can prove this directly
from the definition.
(ii) f(z) = e
z
= e
x
(cos y + i sin y) is entire since
u
x
= e
x
cos y =
v
y
,
u
y
= e
x
sin y =
v
x
.
The derivative is
f
0
(z) =
u
x
+ i
v
x
= e
x
cos y + ie
x
sin y = e
z
,
as expected.
(iii) f
(
z
) =
z
n
for
n N
is entire. This is less straightforward to check. Writing
z = r(cos θ + i sin θ), we obtain
u = r
n
cos , v = r
n
sin .
We can check the Cauchy-Riemann equation using the chain rule, writing
r
=
p
x
2
= y
2
and
tan θ
=
y
x
. This takes quite a while, and it’s not worth
your time. But if you really do so, you will find the derivative to be
nz
n1
.
(iv)
Any rational function, i.e.
f
(
z
) =
P (z)
Q(z)
where
P, Q
are polynomials, is
analytic except at points where
Q
(
z
) = 0 (where it is not even defined).
For instance,
f(z) =
z
z
2
+ 1
is analytic except at ±i.
(v)
Many standard functions can be extended naturally to complex functions
and obey the usual rules for their derivatives. For example,
sin z
=
e
iz
e
iz
2i
is differentiable with derivative
cos z
=
e
iz
+e
iz
2
. We
can also write
sin z = sin(x + iy)
= sin x cos iy + cos x sin iy
= sin x cosh y + i cos x sinh y,
which is sometimes convenient.
Similarly
cos z, sinh z, cosh z
etc. differentiate to what we expect them
to differentiate to.
log z = log |z| + i arg z has derivative
1
z
.
The product rule, quotient rule and chain rule hold in exactly the
same way, which allows us to prove (iii) and (iv) easily.
Examples of non-analytic functions
Example.
(i) Let f(z) = Re z. This has u = x, v = 0. But
u
x
= 1 6= 0 =
v
y
.
So Re z is nowhere analytic.
(ii)
Consider
f
(
z
) =
|z|
. This has
u
=
p
x
2
+ y
2
, v
= 0. This is thus nowhere
analytic.
(iii)
The complex conjugate
f
(
z
) =
¯z
=
z
=
x iy
has
u
=
x, v
=
y
. So the
Cauchy-Riemann equations don’t hold. Hence this is nowhere analytic.
We could have deduced (ii) from this if
|z|
were analytic, then so would
|z|
2
, and hence ¯z =
|z|
2
z
also has to be analytic, which is not true.
(iv)
We have to be a bit more careful with
f
(
z
) =
|z|
2
=
x
2
+
y
2
. The
Cauchy-Riemann equations are satisfied only at the origin. So
f
is only
differentiable at
z
= 0. However, it is not analytic since there is no
neighbourhood of 0 throughout which f is differentiable.