Chapter1
Preliminaries to Complex Analysis
Berret
July 10,2020
Abstract
This chapter is devoted to the exposition of basic preliminary material
which we use extensively throughout of this book. I will help you solve serval
exercises in this chapter, there may be some spelling mistakes or even wrong
methods.
The sweeping development of mathematics during the last two centuries is due in large
part to the introduction of complex numbers; paradoxically, this is based on the seemingly
absurd notion that there are numbers whose squares are negative.
E. Borel, 1952
1
Exercise 1 is very easy, so we skip it and begin with Exercise 2.
Exercise 2. Let <·, ·> denote the usual inner product in R2 . In other words, if Z = (x1 , y1 ) and W = (x2 , y2 ), then
<Z, W > = x1 x2 + y1 y2 .
Similarly, we may define a Hermitian inner product (·, ·) in C by
(z, w) = zw.
The term Hermitian is used to describe the fact that (·, ·) is not symmetric, but
rather satisfies the relation
(z, w) = zw
Show that
<Z, W > =
f or
all
z, w ∈ C.
1
[(z, w) + (w, z)] = Re(z, w),
2
Where we use the usual identification z = x + iy ∈ C with(x, y) ∈ R2 .
Solution 2. Though this is a straightforward calculation, but we have two ways
to solve it.
We suppose that z = z1 + iz2 , w = w1 + iw2 , so
1
1
[(z, w)+(w, z)] = (zw+wz) = Re(zw) = Re ((z1 + iz2 ) (w1 − iw2 )) = z1 w1 +z2 w2
2
2
W ay1
]
1[
1
1
[(z, w)+(w, z)] = (zw+wz) = (z1 +iz2 )(w1 −iw2 )+(z1 −iz2 )(w1 +iw2 ) = z1 w1 +z2 w2
2
2
2
W ay2
Exercise 3. With ω = seiφ , where s ≥ 0 and φ ∈ R, solve the equation
z n = ω in C where n is a natural number. How many solutions are there?
Solution 3. We suppose that z = teiβ , where t ≥ 0 as s ≥ 0 and n is a natural
number. Then we have
tn eniβ = seiφ
1
φ
2πik
1
We easily find that z = s n e n + n , where k = 0, 1, · · · , n − 1 and s n is the real
nth root of s.
In summary, there are n solutions as k has n values.
Exercise 4. Show that it is impossible to define a total ordering on C. In other
words, one cannot find a relation ≻ between complex numbers so that:
(i) For any two complex numbers z,w, one and only one of the following is true:
z ≻ ω, ω ≻ z or z = ω.
2
(ii) For all z1 , z2 , z3 ∈ C the relation z1 ≻ z2 implies z1 + z3 ≻ z2 + z3 .
(iii) Moreover, for all z1 , z2 , z3 ∈ C with z3 ≻ 0, then z1 ≻ z2 implies z1 z3 ≻
z2 z3 .
Solution 4. We can conclude that i = 0. If not, just for a contradiction, that
i ≻ 0, then −1 = i · i ≻ 0 · i = 0. Now we may suppose that 0 ≻ i, but also
similarly, 0 = 0 · i ≻ i · i = −1 . So we must have i = 0. But then for all z ∈ C we
have z · i = z · 0 = 0. So this relation would not give a trivial total ordering.
Exercise 5 and Exercise 6 is not so easy to write down, so I skip them and we
turn to Exercise 7.
Exercise 7. The family of mappings introduced here plays an important role
in complex analysis. These mappings, sometimes called Blaschke factors, will
reappear in various applications in later chapters.
(a) Let z, w be two complex numbers such that zw ̸= 1.Prove that
w−z
<1 if |z| < 1and |w| < 1,
1 − wz
and also that
w−z
= 1 if |z| = 1or |w| = 1.
1 − wz
(b) Prove that for a fixed w in the unit disc D, the mapping
F : z 7→
w−z
1 − wz
satisfies the following conditions:
(i) F maps the unit disc to itself(that is, F : D → D),and is holomorphic.
(ii) F interchanges 0 and w, namely F (0) = w and F (w) = 0.
(iii) |F (z)| = 1 of |z| = 1.
(iv) F : D → D is bijective.
Solution 7.
(a) We have two ways to solve it.
[W ay1]It is a straightforward calculation.
w−z
We suppose that z = z1 + iz2 ,w = w1 + w2 , then to calculate 1−wz
we just need
to calculate
|w−z|
|1−wz| ,
that is,
w1 − z1 + i(w2 − z2 )
1 − w1 z1 − w2 z2 + i(w1 z2 − w2 z1 )
We need to compare it with 1,
that is, to compare w1 − z1 + i(w2 − z2 ) with 1 − w1 z1 − w2 z2 + i(w1 z2 − w2 z1 ),
that is, to compare (w1 − z1 )2 with (1 − w1 z1 − w2 z2 )2 + (w1 z2 − w2 z1 )2 ,
3
2
2
2
2
that is, to compare |w| + |z| − 2(w1 z1 + w2 z2 ) with |w| |z| + 1 − 2(w1 z1 + w2 z2 ),
2
2
2
2
that is, to compare |w| + |z| with |w| |z| + 1.
2
2
When |w| < 1, |z| < 1, (1 − |w| )(1 − |z| ) > 0.
w−z
So we know that 1−wz
<1
2
2
2
2
w−z
When |w| = 1or |z| = 1, |w| + |z| = |w| |z| + 1, so 1−wz
=1
[W ay2]We use the Maximum modulus principle, page 92, Theorem 4.5 to solve it.
Maybe I should retell it.
Theorem 1. (Maximum modulus principle) If f is a non-constant holomorphic function in a region Ω, then f cannot attain a maximum in Ω.
Suppose that [w] < 1 and [z] = 1, then we have
w−z
w−z
=
=1
1 − wz
z−w
w−z
Since z1 = 1. And since that [w] < 1, we see that the function f (z) := 1−wz
is
holomorphic in D. It is easy to find out that it is not a constant, so it satisfies
|f (z)| < 1 as we use the Maximum modulus principle.
(b)
(i) We already show that F (D) ⊂ D, what we need to do is just to prove that
it is holomorphic.
w−(z+h)
w−z
F (z + h) − F (z)
1−w(z+h) − 1−wz
= lim
h→0
h→0
h
h
wwh − h
ww − 1
= lim
=
h→0 [1 − w(h + z)][1 − wz]h
(1 − wz)2
∀z ∈ D, lim
So F (z) is continuous at each point in D, that means F is holomorphic in the unit
disc.
w−0
w−w
(ii) F (0) = 1−w·0
= w, F (w) = 1−ww
=0
(iii) We already explain it in (a) that F (∂D) ⊆ D.
(iv) F ◦ F =
w−z
w− 1−wz
w−z
1− 1−wz
= z, that means F : D → D is a identity mapping, so it
is also a bijection.
Exercise 8 is also very easy, so we skip it and turn to Exercise 9.
Exercise 9. Show that in polar coordinates, the Cauchy-Riemann equations
take the form
∂u
1 ∂v
1 ∂u
∂v
=
and
=−
∂r
r ∂θ
r ∂θ
∂r
Use these equations to show that the logarithm function defined by
logz = logr + iθ
where
4
z = reiθ with − π < θ < π
is holomorphic in the region r > 0 and −π < θ < π.
Solution 9. A straight calculation can solve it.
We know that z = x + iy,x = ρ cos φ,y = ρ sin φ
then
∂u
∂v ∂u
∂v
f (z) = u(x, y) + iv(x, y),
=
,
=−
∂x
∂y ∂y
∂x
so we have
∂u
∂u ∂x ∂u ∂y
∂u
∂u
=
·
+
·
=
· cos φ +
· sin φ
∂ρ
∂x ∂ρ
∂y ∂ρ
∂x
∂y
Similarly, we also have
∂v ∂x
∂v ∂y
∂v
∂v
∂v
=
·
+
·
=
· (−ρ sin φ) +
· ρ cos φ
∂φ
∂x ∂φ ∂y ∂φ
∂x
∂y
so we have
∂u
1 ∂v
=
∂ρ
ρ ∂φ
just continue to calculate,
∂v
∂v ∂x ∂v ∂y
∂v
∂v
=
·
+
·
=
· cos φ +
· sin φ
∂ρ
∂x ∂ρ ∂y ∂ρ
∂x
∂y
∂u
∂u ∂x
∂u ∂y
∂u
∂u
=
·
+
·
=
· (−ρ sin φ) +
· ρ sin φ
∂φ
∂x ∂φ ∂y ∂φ
∂x
∂y
so we have
1 ∂u
∂v
=−
∂ρ
ρ ∂φ
which means
∂u
1 ∂v 1 ∂u
∂v
=
,
=−
∂r
r ∂θ r ∂θ
∂r
And we check it as followed,
∂u
1 ∂v
= ,
=1
∂r
r ∂θ
so f (z) = logz
∂u
1 ∂v
= ·
∂r
r ∂θ
∂u
∂v
= 0,
=0
∂θ
∂r
1 ∂u
∂v
=−
r ∂θ
∂r
is holomorphic in the region r > 0 and −π < θ < π
Exercise 10. Show that
4
∂ ∂
∂ ∂
=4
= △,
∂z ∂z
∂z ∂z
5
where △ is the Laplacian
△=
∂2
∂2
+ 2.
2
∂x
∂y
Solution 10. To prove it, we shall begin with
∂
∂ ∂z
∂ ∂z
∂
∂
=
·
+
·
=
+
∂x
∂z ∂x ∂z ∂x
∂z
∂z
∂
∂ ∂z
∂
∂ ∂z
∂
∂
=
·
+
·
·
=i
−i
∂y
∂z ∂y ∂z ∂z ∂y
∂z
∂z
∂
∂2
∂2
∂2
∂2
= 2+
+ 2+
2
∂x
∂z
∂z∂z
∂z∂z
∂z
∂2
∂2
∂2
∂2
∂
=− 2 +
− 2+
2
∂y
∂z
∂z∂z
∂z∂z
∂z
△=
∂
∂2
∂
∂2
+
2
+
=
2
∂x2
∂y 2
∂z∂z
∂z∂z
so lastly we get
△=4
∂ ∂
∂ ∂
=4
∂z ∂z
∂z ∂z
Exercise 11. Use Exercise 10 to prove that if f is holomorphic in the open set Ω,
then the real and imaginary parts of f are harmonic; that is, their Laplacian is
zero.
Solution 11.
( 2
)
( )
(
)
∂
∂2
∂ ∂v
∂
∂v
∂2v
∂2v
△u =
+
u
=
+
−
=
−
∂x2
∂y 2
∂x ∂y
∂y
∂x
∂x∂y ∂y∂x
( 2
)
(
)
( )
∂
∂2
∂
∂u
∂ ∂u
∂2v
∂2v
△v =
+
v
=
−
+
=−
+
2
2
∂x
∂y
∂x
∂y
∂y ∂x
∂x∂y ∂y∂x
f = u + iv is a holomorphic function and u, v is continuous in the defined area.
Thus,
∂2v
∂2u
∂2u
∂2u
=
,
=
∂x∂y
∂y∂x
∂x∂y
∂y∂x
So the real and imaginary parts of f is harmonic; that is, their Laplacian: △u, △v
is zero.
Exercise 12. Consider the function defined by
√
f (x + iy) = |x| |y|, where x, y ∈ R.
Show that f satisfies the Cauchy-Riemann equations at the origin, yet f is not
holomorphic at 0.
6
√
Solution 12. We can easily get that u = |x| |y| and v = 0
v is continuously differentiable while u is differentiable but not continuously differentiable.
∂u
∂v
∂v
At the origin, ∂u
∂x = ∂y = 0, and as v = 0, ∂x = ∂y = 0, which means that f
satisfies the Cauchy-Riemann equations at the origin, yet f is not holomorphic at 0.
Exercise 13. Suppose that f is holomorphic in an open set Ω. Prove that in
any one of the following cases:
(i) Re (f ) is constant
(ii) Im (f ) is constant
(iii) |f | is constant
one can conclude that f is constant.
Solution 13. Let f (z) = f (x, y) = u (x, y) + iv (x, y), where z = x + iy.
(i) Since Re (f ) is constant,
∂u
∂u
=
=0
∂x
∂y
as we know that f is holomorphic in an open set Ω, by the Cauchy-Riemann
equations,
∂v
∂u
=−
=0
∂x
∂y
then in Ω,
∂f
∂u
∂v
f ′ (z) =
=
+i
=0+0=0
∂x
∂x
∂x
thus f (z) is constant.
(ii) Since Im (f ) is constant,
∂v
∂v
=
=0
∂x
∂y
by the Cauchy-Riemann equations,
∂u
∂v
=−
=0
∂x
∂y
Thus, in Ω,
f ′ (z) =
∂f
∂u
∂v
=
+i
=0+0=0
∂x
∂x
∂x
thus f (z) is constant.
(iii) We first give a mostly correct
√ argument; the reader should pay attention
to find the difficulty. Since |f | = x2 + y 2 is constant, we have
∂(u2 + v 2 )
∂u
∂v
= 2u
+ 2v
∂x
∂x
∂x
∂(u2 + v 2 )
∂u
∂v
0=
= 2u
+ 2v
∂x
∂y
∂y
0=
7
Plug in the Cauchy-Riemann equations and we get
∂v
∂v
+v
=0
∂y
∂x
∂v
∂v
−u
+v
=0
∂x
∂y
u
thus we have
∂v
v ∂v
=
∂x
u ∂y
u2 + v 2 ∂v
=0
u
∂y
∂v
which means u2 + v 2 = 0 or ∂y
= 0. If u2 + v 2 = 0, then, since u, v are real,
u = v = 0, and thus f = 0 which is constant. Thus we may assume u2 + v 2 equal
a non-zero constant, and we may divide by it. We multiply both sides by u and
∂v
∂v
find ∂y
= 0, plug back in the early equation, we get ∂x
= 0, by Cauchy-Riemann
∂u
equations, ∂x = 0
∂f
∂u
∂v
f ′ (z) =
=
+i
=0
∂x
∂x
∂x
Thus f is constant.
Why is the above only mostly a proof? The problem is we have a division by
u, and need to make sure everything is well-defined. Specifically, we need to know
that u is never zero. We do have f ′ = 0 except at points where u = 0, but we
would need to investigate that a bit more.
Let’s turn back to
∂(u2 + v 2 )
∂u
∂v
= 2u
+ 2v
∂x
∂x
∂x
∂(u2 + v 2 )
∂u
∂v
0=
= 2u
+ 2v
∂x
∂y
∂y
0=
Plug in the Cauchy-Riemann equations and we get
∂v
∂v
+v
=0
∂y
∂x
∂v
∂v
+v
=0
−u
∂x
∂y
u
We multiply the first equation u and the second by v, and obtain
∂v
∂v
+ uv
=0
∂y
∂x
∂v
∂v
−uv
+ v2
=0
∂x
∂y
u2
Adding the two yields
u2
∂v
∂v
+ v2
=0
∂y
∂y
8
or equivalently
∂v
=0
∂y
We now argue in a similar manner as before, except now we don’t have the annoying u in the denominator. If u2 + v 2 = 0 then u = v = 0, else we can divide by
∂v
u2 + v 2 = 0 and find ∂y
= 0. Arguing along these lines finishes the proof.
One additional remark by Jeff Meng: we can trivially pass from results
on patrials with respect to v to those with respect to u by noting that if f = u + iv
has constant magnitude, so too does g = if = −v + iu, which essentially switches
the roles of u and v, Though this isn’t needed for this problem, arguments such
as this can be very useful.
The following is from Steven Miller. Let’s consider the proof. If |f | = 0
the problem is trivial as then f = 0, so we assume |f | equals a non-zero constant.
As |f | is constant, |f |2 = f f is constant. By the quotient rule, the ratio of two
holomorphic functions is holomorphic, assuming the denominator is non-zero. We
2
thus find f = |ff| is holomorphic. Thus f and f are holomorphic, and satisfy the
Cauchy-Riemann equations. Applying these to f = ui + iv yields
(u2 + v 2 )
∂v ∂u
∂v
∂u
=
,
=−
∂x
∂y ∂y
∂x
while applying to f = u + i(−v) gives
∂u
∂(−v) ∂u
∂(−v)
=
,
=−
∂x
∂y
∂y
∂x
Adding these equations together yields
2
∂u
∂u
= 0, 2
=0
∂x
∂y
Thus u is constant, and by (i) implies that f is constant. If we didn’t want to use
that we could subtract rather than add, and similarly find that v is constant.
Exercise 14 and Exercise 15 is the proof of summations by parts and Abel’s
theorem, the solution is similar to Real Analysis, so we skip and just retell them.
N
N
Theorem 2. (summations by parts) Suppose {an }n=1 and {bn }n=1 are two
∑k
finite sequences
∑ of complex numbers. Let Bk = n=1 bn denote the partial sums
of the series
bn with the convention B0 = 0, then
N
∑
an bn = aN bN − aM bM −
n=M
N
−1
∑
(an+1 − an )Bn
n=M
Theorem 3. (Abel’s theorem) Suppose
lim
r→1,r<1
∞
∑
∑∞
n=1
rn an =
n=1
an converges, then
∞
∑
n=1
9
an
As to Exercise 16,17, we use Hadamard’s formula and directly solve them. Actually, we can use Real Analysis to solve them, so I skip them.
Exercise 18. Let f be a power series centered at the origin. Prove that f
has a power series expansion around any point in its disc of convergence.
Solution 18. We suppose
+∞
∑
f (z) =
an z n
n=0
and for any z0 ∈ Ω, then write z = z0 + (z − z0 ), as we know that f (z0 ) converges,
we have
f (z) =
=
=
+∞
∑
n=0
+∞
∑
n=0
+∞
∑
an [z0 + (z − z0 )]n
an [
n
∑
Cni z0n−i (z − z0 )i ]n
i=0
bn (z − z0 )n
n=0
where
+∞
∑
bn =
n
Cn+i
an+i z0i
i=0
and bn actually also converges.
Exercise 19. Prove the following:
∑
(i) The power series ∑nz n does not converge on any point of the unit circle.
(ii) The power series ∑ z n /n2 converges at any point of the unit circle.
(iii) The power series
z n /n converges at every point of the unit circle except
z = 1.
Solution 19.
(i) On the unit circle, we suppose z = cos θ + i sin θ, θ ∈ [0, 2π), then an =
nz n = n cos θ + in sin θ
We know that limn→∞ n cos nθ doesn’t exist ∀θ ∈ (0, 2π). But when θ = 0, we
have limn→∞ n cos nθ = +∞, ∑
so limn→+∞ an doesn’t exist too.
∑ n
+∞
Now we can deduce that n=1 an diverges, that is, The power series
nz
does not converge on any point of the unit circle.
(ii) On the unit circle, we suppose z = cos θ + i sin θ, θ ∈ [0, 2π), then αn =
z n /n2 = cos nθ/n2 + i sin θ/n2 = an + ibn
∞
∑
an =
n=1
As
∑∞
1
n=1 n2
converges, so
∞
∞
∑
∑
cos nθ
1
≤
2
2
n
n
n=1
n=1
∑∞
n=1
an converges.
10
Similarly,
∞
∑
n=1
∑∞
bn =
∞
∞
∑
∑
sin nθ
1
≤
2
2
n
n
n=1
n=1
∑∞
As n=1 n12 ∑
converges, so n=1 bn converges.
∞
In summary, n=1 α also converges as to be proved
above.
(iii) On the unit circle, we suppose z = cos θ + i sin θ, θ ∈ [0, 2π), then
αn = z n /n = cosnnθ + i sinn nθ = an + ibn
∑∞
∑+∞ n
When θ = 0, that is, when z = 1, n=1 zn = n=1 n1 obviously diverges.
∑ cos nθ
∑ sin nθ
When θ ∈ (0, 2π),
and
both converges by Dirichlet disn
n
criminance.
For the integrity of the proof, I will explain it briefly.
∑
sin(n+ 1 )θ
As can be easily proved, cos nθ’s partial sum sequence is 2 sin θ2 − 12 , which
2
{ }
is bounded when θ ∈ (0, 2π), n1 is monotonically decreasing and tending to zero,
∑ cos nθ
∑
so
converges. Similarly, ∑ sinnnθ also converges.
n
In summary, the power series z n /n converges at every point of the unit circle
except z = 1.
Exercise 20. Expand (1 − z)−m in powers of z. Here m is a fixed positive
integer. Also, show that if
(1 − z)−m =
∞
∑
an z n
n=0
then one obtains the following asymptotic relation for the coefficients:
an ∼
1
nm−1
(m − 1)!
as n → ∞
Solution 20. Directly use Taylor’s formula at zero, we get
(1 − z)−m =
∞
∑
an z n
n=0
where
an =
(m + n − 1)!
(m − 1)!n!
To prove
an ∼
1
nm−1
(m − 1)!
as n → ∞
we just need to prove
(m + n − 1)!
1
∼
nm−1
(m − 1)!n!
(m − 1)!
as n → ∞
that is, to prove
(n + 1) · · · (m + n − 1) ∼ nm−1
11
The number of items in the left formula is actually m − 1, thus we just need to
prove
1
2
m−1
)∼1
(1 + )(1 + ) · · · (1 +
n
n
n
We already learn that m is a fixed positive integer and n → ∞, thus the number
of items in the left formula is limited, we can directly prove it.
Exercise 21. Show that for |z| < 1, one has
n
z
z2
z2
z
+
+
·
·
·
+
n+1 + · · · =
2
4
2
1−z
1−z
1−z
1−z
and
k
2z 2
2k z 2
z
z
+
+
·
·
·
+
+ ··· =
1+z
1 + z2
1−z
1 + z 2k
Justify any change in the order of summation.
Solution 21. For the first formula, a calculation can solve it.
n
−
z
z
z2
z2
+
+
+ ··· +
+ ···
2
4
1−z
1−z
1−z
1 − z 2n+1
n
z2
z2
−z 2
+
+
·
·
·
+
+ ···
=
1 − z2
1 − z4
1 − z 2n+1
n
=
−z 4
z2
+ ···
+ +··· +
4
1−z
1 − z 2n+1
···
n
z2
= lim
n
n→∞ 1 − z 2
for |z| < 1, the formula equals 0, which means
n
z
z
z2
z2
+ ··· =
+
+ ··· +
2
4
1−z
1−z
1−z
1 − z 2n+1
As for the second formula, similarly,
k
−
z
z
2z 2
2k z 2
+ ···
+
+
+ ··· +
2
1−z
1+z
1+z
1 + z 2k
k
=
−2z 2
2z 2
2k z 2
+
+ ··· +
+ ···
2
2
1−z
1+z
1 + z 2k
k
=
−22 z 4
2k z 2
+
·
·
·
+
+ ···
1 − z4
1 + z 2k
···
n
−2n z 2
= lim
n
n→∞ 1 + z 2
12
for |z| < 1, the formula equals 0, which means
k
2z 2
2k z 2
z
z
+
+
·
·
·
+
k + ··· =
2
2
1+z
1+z
1−z
1+z
There must be some concise solutions, but the solution here is very easy to come
out.
Exercise 22. Let N = {1, 2, 3, · · · } denotes the set of positive integers. A subset
S ⊂ N is said to be in arithmetic progression if
S = {a, a + d, a + 2d, a + 3d, · · · }
where a, d ∈ N. Here d is called the step of S.
Show that N cannot be partitioned into a finite number of subsets that are in
arithmetic progression with distinct steps (except for the trivial case a = d = 1).
Solution 22. Suppose Sa,d = {a, a + d, a + 2d, a + 3d, · · · }
∑
)
za (
(n+1)d
1
−
z
n→∞ 1 − z d
z n = z a + z a+d + z a+2d + · · · + z a+nd + · · · = lim
n∈Sa,d
let |z| < 1, we have
∑
zn =
za
1 − zd
zn =
z
1−z
n∈Sa,d
at the same time, we have
∑
n∈N∗
To find a finite number of subsets with different elements and steps d,we can just
pick by
1, 3, 5, 7, 9 · · · ; 2, 6, 10, · · · ; · · ·
so we can get
Sn =
∑
∑
zn +
n∈Sa1 ,d1
n∈Sa2 ,d2
zn
n∈San ,dn
2n
2
=
∑
zn + · · · +
z
z
z
+
+ ··· +
2
4
1−z
1−z
1 − z 2n+1
then
n−1
Sn +
n
1
z
z2
z2
1 + z2
+
+ ··· +
n +
n+1 =
2
4
2
P
2
1−z
1−z
1−z
1−z
1 − z 2n+1
1
z
−
1−z
1 − z 2n+1
z
∴ lim Sn =
n→∞
1−z
∴ Sn =
13
To any subsets with finite numbers, such as Sa1 ,d1 , Sa2 ,d2 , · · · , San ,dn
∑
Sai ,di < lim Sn =
n→∞
z
1−z
So to subsets with
if a and d cannot be 1 at the same time, it’s
∑ finite numbers,
z
impossible that
Sai ,di = 1−z
.Q.E.D.
Exercise 23. Consider the function f defined on R by
{
0
if x ≤ 0
f (x) =
−1/x2
e
if x > 0
Prove that f is indefinitely differentiable on R, and that f (n) (0) = 0 for
n ≥ 1.
∑all
∞
Conclude that f does not have a converging power series expansion n=0 an xn
for x near the origin.
Solution 23. When x ∈ (−∞, 0], f (0) (x) = 0 is obvious. So it’s derivable of any
order and
lim− f (n) (x) = 0
x→0
− x2
When x ∈ (0, +∞), n = 1, f ′ (x) = −2
Use mathematical methods of
x3 e
induction, then suppose when n = k,
1
f (k) (x) = e− x2 · P (x),
1
P (x) =
∑
Ci
1
xi
then when n = k + 1, we can deduce that
1
2
f (k+1) (x) = e− x2 · P ′ (x) − 3
x
[
]
2
− x12
′
=e
· P (x) − 3 P (x)
x
= e− x2 · Q(x)
1
where
Q(x) =
and
∑
Ci′
1
xi
e− x2
=0
n→∞ xn
1
lim
Obviously, f (x)’s derivable of any order of k can be expressed as e− x2 ·P (x), where
P (x) is a polynomial function of x. And when x ∈ (0, +∞), P (x) is derivable
everywhere, so f (x) can be derived at any order.∑
∞
1
Then in a small neighborhood (0, δ), f (x) = n=0 n!
· (− x12 )n .
As to
1
1
lim
· (− 2 )n
n→∞,x→0 n!
x
1
14
∀n, M > 0, we can find x, when x satisfies
{ √
2n
x < min
we have
1
,δ
n!M
}
1
1
· (− 2 )n > M
n!
x
so the series does not converge near the origin.
Exercise 25. The next three calculations provide some insight into Cauchys
theorem, which we treat in the next chapter.
(i) Evaluate the integrals
∫
z n dz
γ
for all integers n. Here γ is any circle centered at the origin with the positive
(counterclockwise) orientation.
(ii) Same question as before, but with γ any circle not containing the origin.
(iii) Show that if |a| < r < |b|, then
∫
2πi
1
dz =
(z
−
a)(z
−
b)
a
−b
γ
where γ denotes the circle centered at the origin, of radius r, with the positive
orientation.
Solution 25.
(i) Suppose z = reiθ , θ ∈ [0, 2π), then we have
∫
∫
z n dz =
rn eiθn dreiθ
γ
γ
=r
∫
n+1
2π
ieiθ(n+1) dθ
0
∫ 2π
rn+1
=
·
dei(n+1)θ
n+1 0
]
rn+1 [ i(n+1)2π
=
· e
−1
n+1
=0
(ii) Suppose z = reiθ + keiφ , θ, φ ∈ [0, 2π). φ, r, k here are all constant.
15
Then
∫
z n dz =
γ
z n+1
n+1
=
γ
2π
(reiθ + keiφ )n+1
n+1
0
(rei2π + keiφ )n+1
(r + keiφ )n+1
=
−
n+1
n+1
iφ n+1
iφ n+1
(r + ke )
(r + ke )
−
=
n+1
n+1
=0
(iii)
∫
γ
∫ (
1
1
dz =
(z − a)(z − b)
a−b
equals
∫
γ
let
1
dz −
z−a
γ
∫
)
1
− f rac1z − b dz
z−a
1
dz = 2πi
z−b
γ
z ′ = z − a = reiθ − a = r′ eiθ
∫
γ
1
d(z − a) =
z−a
∫
γ
∫
′
1 ′
dz
z′
=
′ iθ ′
dθ′
γ ir′ eiθ′
r e
2π+arcsin
r−a
r′
= iθ
arcsin
r−a
r′
= 2πi
r−a
r−a
′
here θ from 0 to 2π,
∫ θ 1 from arcsin r′ to 2π + arcsin r′
We know that γ z−b dz is analytic,
∫
∴
γ
In summary,
∫
γ
1
1
dz =
(z − a)(z − b)
a−b
1
dz = 0
z−b
∫ (
γ
)
1
2πi
− f rac1z − b dz =
z−a
a−b
Q.E.D. Or we have another
∫ 1concrete proof, I’ll show you behind.
dz.
Firstly we calculate γ z−a
∫
γ
1
dz =
z−a
∫
γ
d(z − a)
=
z−a
16
∫
C
′
rm d(eiθ ) + d(rm )·iθ
rm eiθ′
′
z − a = rm eiθ
′
2
rm
= ra2 + r2 − 2rra cos(θ − θa )
thus we have
∫
=
=
=
=
=
′
rra sin(θ−θa )+rm ·eiθ
dθ
rm
′
iθ
rm e
C
∫ 2π
∫ 2π
sin(θ − θa )
idθ + rra
dθ
2
rm
0
0
∫ 2π
sin(θ − θa )
dθ
2πi + rra
2
2
ra + r − 2rra cos(θ − θa )
0
∫ 2π
d[ra2 + r2 − 2rra cos(θ − θa )]
2πi + 2
ra2 + r2 − 2rra cos(θ − θa )
0
2π
2πi + 2ln ra2 + r2 − 2rra cos(θ − θa )
0
′
irm eiθ dθ′ +
= 2πi
∫
Similarly, we calculate γ
1
z−b dz
then, and we get
∫
γ
1
dz = 0
z−b
Exercise 26.Suppose f is continuous in a region Ω. Prove that any two primitives
of f (if they exist) differ by a constant.
Solution 26. We suppose F, G are any two primitive functions of f . Suppose
F = F − G, then
[H]′ = [F − G]′ = F ′ − G′ = f − f = 0
we may suppose H = u + iv, because f is continuous, so F, G is analytic, therefore
H is also analytic.
∂u
∂v
∴
+i
=0
∂x
∂x
by Cauchy-Riemann equations,
∂u
∂v
=
,
∂x
∂y
∴
∂u
∂v
=
∂y
∂x
∂u
∂v
=
=0
∂y
∂x
∴ u = c1 , v = c2 ,
c1 , c 2
are constants.
∴ F − G = c1 + ic2
17
Q.E.D.
ω=
i
x
y
i
=
= 2
·i+ 2
2
z
x + iy
x +y
x + y2
ω = x′ + iy ′
by comparison,
y
+ y2
x
y′ = 2
x + y2
x′ =
x2
And
x2 + y 2 − 2y = 0
we get
x′ =
18
1
2