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Chapter 1
Complex integration
1.1 Complex number quiz
- Simplify (^) 3+4^1 i.
- Simplify | (^) 3+4^1 i |.
- Find the cube roots of 1.
- Here are some identities for complex conjugate. Which ones need correc- tion? z + w = ¯z + ¯w, z − w = ¯z − w¯, zw = ¯z w¯, z/w = ¯z/ w¯. Make suitable corrections, perhaps changing an equality to an inequality.
- Here are some identities for absolute value. Which ones need correction? |z + w| = |z| + |w|, |z − w| = |z| − |w|, |zw| = |z||w|, |z/w| = |z|/|w|. Make suitable corrections, perhaps changing an equality to an inequality.
- Define log(z) so that −π < = log(z) ≤ π. Discuss the identities elog(z)^ = z and log(ew) = w.
- Define zw^ = ew^ log^ z^. Find ii.
- What is the power series of log(1 + z) about z = 0? What is the radius of convergence of this power series?
- What is the power series of cos(z) about z = 0? What is its radius of convergence?
- Fix w. How many solutions are there of cos(z) = w with −π < <z ≤ π.
2 CHAPTER 1. COMPLEX INTEGRATION
1.2 Complex functions
1.2.1 Closed and exact forms
In the following a region will refer to an open subset of the plane. A differential form p dx + q dy is said to be closed in a region R if throughout the region
∂q ∂x
∂p ∂y
It is said to be exact in a region R if there is a function h defined on the region with
dh = p dx + q dy. (1.2)
Theorem. An exact form is closed. The converse is not true. Consider, for instance, the plane minus the origin. The form (−y dx + x dy)/(x^2 + y^2 ) is not exact in this region. It is, however, exact in the plane minus the negative axis. In this region
−y dx + x dy x^2 + y^2
= dθ, (1.3)
where −π/ 2 < θ < π/2. Green’s theorem. If S is a bounded region with oriented boundary ∂S, then ∫
∂S
p dx + q dy =
S
∂q ∂x
∂p ∂y
) dx dy. (1.4)
Consider a region R and an oriented curve C in R. Then C ∼ 0 (C is homologous to 0) in R means that there is a bounded region S such that S and its oriented boundary ∂S are contained in R such that ∂S = C. Corollary. If p dx + q dy is closed in some region R, and if C ∼ 0 in R, then ∫
C
p dx + q dy = 0. (1.5)
If C is an oriented curve, then −C is the oriented curve with the opposite orientation. The sum C 1 + C 2 of two oriented curves is obtained by following one curve and then the other. The difference C 1 − C 2 is defined by following one curve and then the other in the reverse direction. Consider a region R and two oriented curves C 1 and C 2 in R. Then C 1 ∼ C 2 (C 1 is homologous to C 2 ) in R means that C 1 − C 2 ∼ 0 in R. Corollary. If p dx + q dy is closed in some region R, and if C 1 ∼ C 2 in R, then (^) ∫
C 1
p dx + q dy =
C 2
p dx + q dy. (1.6)
4 CHAPTER 1. COMPLEX INTEGRATION
Example: Consider the differential form zm^ dz for integer m 6 = 1. When m ≥ 0 this is defined in the entire complex plane; when m < 0 it is defined in the punctured plane (the plane with 0 removed). It is exact, since
zm^ dz =
m + 1
dzm+1. (1.17)
On the other hand, the differential form dz/z is closed but not exact in the punctured plane.
1.2.4 Polar representation
The exponential function is defined by
exp(z) = ez^ =
∑^ ∞
n=
zn n!
It is easy to check that
ex+iy^ = exeiy^ = ex(cos(y) + i sin(y)). (1.19)
Sometimes it is useful to represent a complex number in the polar represen- tation z = x + iy = r(cos(θ) + i sin(θ)). (1.20)
This can also be written z = reiθ^. (1.21) From this we derive
dz = dx + i dy = dr eiθ^ + rieiθ^ dθ. (1.22)
This may also be written dz z
dr r
Notice that this does not say that dz/z is exact in the punctured plane. The reason is that the angle θ is not defined in this region. However dz/z is exact in a cut plane, that is, a plane that excludes some line running from the origin to infinity. Let C(0) be a circle of radius r centered at 0. We conclude that ∫
C(0)
f (z) dz =
∫ (^2) π
0
f (z)z i dθ. (1.24)
In particular, (^) ∫
C(0)
z
dz =
∫ (^2) π
0
i dθ = 2πi. (1.25)
By a change of variable, we conclude that for a circle C(z) of radius r centered at z we have (^) ∫
C(z)
ξ − z
dξ = 2πi. (1.26)
1.3. COMPLEX INTEGRATION AND RESIDUE CALCULUS 5
1.2.5 Branch cuts
Remember that dz z
dr r
is exact in a cut plane. Therefore
dz z = d log(z) (1.28)
in a cut plane, log(z) = log(r) + iθ (1.29)
Two convenient choices are 0 < θ < 2 π (cut along the positive axis and −π < θ < π (cut along the negative axis). In the same way one can define such functions as √ z = exp(
log(z)). (1.30)
Again one must make a convention about the cut.
1.3 Complex integration and residue calculus
1.3.1 The Cauchy integral formula
Theorem. (Cauchy integral formula) Let f (ξ) be analytic in a region R. Let C ∼ 0 in R, so that C = ∂S, where S is a bounded region contained in R. Let z be a point in S. Then
f (z) =
2 πi
C
f (ξ) ξ − z
dξ. (1.31)
Proof: Let Cδ (z) be a small circle about z. Let R′^ be the region R with the point z removed. Then C ∼ Cδ (z) in R′. It follows that
1 2 πi
C
f (ξ) ξ − z
dξ =
2 πi
Cδ (z)
f (ξ) ξ − z
dξ. (1.32)
It follows that
1 2 πi
C
f (ξ) ξ − z
dξ − f (z) =
2 πi
Cδ (z)
f (ξ) − f (z) ξ − z
dξ. (1.33)
Consider an arbitrary ² > 0. The function f (ξ) is continuous at ξ = z. Therefore there is a δ so small that for ξ on Cδ (z) the absolute value |f (ξ)−f (z)| ≤ ². Then the integral on the right hand side has integral with absolute value bounded by
1 2 π
∫ (^2) π
0
δ
δdθ = ². (1.34)
Therefore the left hand side has absolute value bounded by ². Since ² is arbitrary, the left hand side is zero.
1.4. PROBLEMS 7
1.3.4 A residue calculation
Consider the task of computing the integral ∫ (^) ∞
−∞
e−ikx^
x^2 + 1
dx (1.42)
where k is real. This is the Fourier transform of a function that is in L^2 and also in L^1. The idea is to use the analytic function
f (z) = e−ikz^
z^2 + 1
The first thing is to analyze the singularities of this function. There are poles at z = ±i. Furthermore, there is an essential singularity at z = ∞. First look at the case when k ≤ 0. The essential singularity of the exponen- tial function has the remarkable feature that for =z ≥ 0 the absolute value of e−ikz^ is bounded by one. This suggests looking at a closed oriented curve Ca in the upper half plane. Take Ca to run along the x axis from −a to a and then along the semicircle z = aeiθ^ from θ = 0 to θ = π. If a > 1 there is a singularity at z = i inside the curve. So the residue is the value of g(z) = e−ikz^ /(z + i) at z = i, that is, g(i) = ek/(2i). By the residue theorem ∫
Ca
e−ikz^
z^2 + 1
dz = πek^ (1.44)
for each a > 1. Now let a → ∞. The contribution from the semicircle is bounded by ∫ (^) π
0
a^2 − 1
a dθ = π
a a^2 − 1
We conclude that for k ≤ 0 ∫ (^) ∞
−∞
e−ikx^
x^2 + 1
dx = πek. (1.46)
Next look at the case when k ≥ 0. In this case we could look at an oriented curve in the lower half plane. The integral runs from a to −a and then around a semicircle in the lower half plane. The residue at z = −i is e−k/(− 2 i). By the residue theorem, the integral is −πe−k. We conclude that for all real k ∫ (^) ∞
−∞
e−ikx^
x^2 + 1
dx = πe−|k|. (1.47)
1.4 Problems
- Evaluate (^) ∫ (^) ∞
0
x(4 + x^2 )
dx
by contour integration. Show all steps, including estimation of integrals that vanish in the limit of large contours.
8 CHAPTER 1. COMPLEX INTEGRATION
- In the following problems f (z) is analytic in some region. We say that f (z) has a root of multiplicity m at z 0 if f (z) = (z − z 0 )mh(z), where h(z) is analytic with h(z 0 ) 6 = 0. Find the residue of f ′(z)/f (z) at such a z 0.
- Say that f (z) has several roots inside the contour C. Evaluate
1 2 πi
C
f ′(z) f (z)
dz.
- Say that f (z) = a 0 + a 1 z + a 2 z^2 + · · · + anzn
is a polynomial. Furthermore, suppose that C is a contour surrounding the origin on which |akzk| > |f (z) − akzk|.
Show that on this contour
f (z) = akzkg(z)
where |g(z) − 1 | < 1
on the contour. Use the result of the previous problem to show that the number of roots (counting multiplicity) inside C is k.
- Find the number of roots (counting multiplicity) of z^6 + 3z^5 + 1 inside the unit circle.
1.5 More residue calculus
1.5.1 Jordan’s lemma
Jordan’s lemma says that for b > 0 we have
1 π
∫ (^) π
0
e−b^ sin(θ)^ dθ ≤
b
To prove it, it is sufficient to estimate twice the integral over the interval from 0 to π/2. On this interval use the inequality (2/π)θ ≤ sin(θ). This gives
π
∫ π 2
0
e−^2 bθ/π^ dθ =
b
(1 − e−b) ≤
b
10 CHAPTER 1. COMPLEX INTEGRATION
1.5.4 Poles of higher order
If f (z) = g(z)/(z − z 0 )m^ with g(z) analytic near z 0 and g(z 0 ) 6 = 0 and m ≥ 1, then f (z) is said to have a pole of order m at z 0. Theorem. If f (z) = g(z)/(z − z 0 )m^ has a pole of order m, then its residue at that pole is
res(z 0 ) =
(m − 1)!
g(m−1)(z 0 ). (1.57)
Proof. By the Cauchy formula for derivatives for each sufficiently small circle C about z the m − 1th derivative satisfies
1 (m − 1)! g(m−1)(z 0 ) =
2 πi
C
g(z) (z − z 0 )m^ dz =
2 πi
C
f (z) dz. (1.58)
The expression given in the theorem is evaluated in practice by using the fact that g(z) = (z − z 0 )mf (z) for z near z 0 , performing the differentiation, and then setting z = z 0. This is routine, but it can be tedious.
1.5.5 A residue calculation with a double pole
Confider the task of computing the integral ∫ (^) ∞
−∞
e−ikx^
(x^2 + 1)^2 dx (1.59)
where k is real. This is the Fourier transform of a function that is in L^2 and also in L^1. The idea is to use the analytic function
f (z) = e−ikz^
(z^2 + 1)^2
The first thing is to analyze the singularities of this function. There are poles at z = ±i. Furthermore, there is an essential singularity at z = ∞. First look at the case when k ≤ 0. Consider a closed oriented curve Ca in the upper half plane. Take Ca to run along the x axis from −a to a and then along the semicircle z = aeiθ^ from θ = 0 to θ = π. If a > 1 there is a singularity at z = i inside the curve. The pole there is of order 2. So the residue is calculated by letting g(z) = e−ikz^ /(z + i)^2 , taking the derivative, and evaluating at z = i, that is, g′(i) = (1 − k)ek/(4i). By the residue theorem ∫
Ca
e−ikz^
z^2 + 1
dz =
π(1 − k)ek^ (1.61)
for each a > 1. Now let a → ∞. The contribution from the semicircle vanishes in that limit. We conclude that for k ≤ 0 ∫ (^) ∞
−∞
e−ikx^
(x^2 + 1)^2
dx =
π(1 − k)ek. (1.62)
1.6. THE TAYLOR EXPANSION 11
Next look at the case when k ≥ 0. In this case we could look at an oriented curve in the lower half plane. The integral runs from a to −a and then around a semicircle in the lower half plane. The residue at z = −i is −(1 + k)e−k/(4i). By the residue theorem, the integral is −π(1 + k)e−k/2. We conclude that for all real k (^) ∫ (^) ∞
−∞
e−ikx^
(x^2 + 1)^2
dx =
π(1 + |k|)e−|k|^ (1.63)
1.6 The Taylor expansion
1.6.1 Radius of convergence
Theorem. Let f (z) be analytic in a region R including a point z 0. Let C(z 0 ) be a circle centered at z 0 such that C(z 0 ) and its interior are in R. Then for z in the interior of C(z 0 )
f (z) =
∑^ ∞
m=
m!
f (m)(z 0 )(z − z 0 )m. (1.64)
Proof. For each fixed ξ the function of z given by
1 ξ − z
ξ − z 0 + z 0 − z
(ξ − z 0 )
1 − z ξ−−zz^00
(ξ − z 0 )
∑^ ∞
m=
z − z 0 ξ − z 0
)m .
(1.65) has a geometric series expansion. Multiply by (1/ 2 πi)f (ξ) dξ and integrate around C(z 0 ). On the left hand side apply the Cauchy integral formula to get f (z). In each term in the expansion on the right hand side apply the Cauchy formula for derivatives in the form
1 2 πi
C(z 0 )
f (ξ) (ξ − z 0 )m+^
dξ =
m!
f (m)(z 0 ). (1.66)
This theorem is remarkable because it shows that the condition of analyticity implies that the Taylor series always converges. Furthermore, take the radius of the circle C(z 0 ) as large as possible. The only constraint is that there must be a function that is analytic inside the circle and that extends f (z). Thus one must avoid the singularity of this extension of f (z) that is closest to z 0. This explains why the radius of convergence of the series is the distance from z 0 to this nearest singularity. The reason that we talk of an analytic extension is that artificialities in the definition of the function, such as branch cuts, should not matter. On the other hand, singularities such as poles, essential singularities, and branch points are intrinsic. The radius of convergence is the distance to the nearest such intrinsic singularity. If one knows an analytic function near some point, then one knows it all the way out to the radius of convergence of the Taylor series about that point.
1.6. THE TAYLOR EXPANSION 13
value of the logarithm function on the point with coordinates z, w on the curve exp(w) = z is w. Notice that if z is given, then there are infinitely many corresponding values of w. Thus if we want to think of log z as a function of a complex variable, then it is ambiguous. But if we think of it as a function on the Riemann surface, then it is perfectly well defined. The Riemann surface has infinitely many sheets. If we wish, we may think of each sheet as the complex plane cut along the negative axis. As z crosses the cut, it changes from one sheet to the other. The value of w also varies continuously, and it keeps track of what sheet the z is on. The infinitely many sheets form a kind of spiral that winds around the origin infinitely many times.
