"G12CAN Complex Analysis Books: Schaum Outline book on Complex Variables (by M. Spiegel), or Churchill and Brown, Complex Analysis and Applications. There should be copies in Short Loan and Reference Only sections of the library. Notes are on www.maths.nottingham.ac.uk/personal/jkl (readable in PDF form). Lecturer: J.K. Langley (C121, jkl@maths, (95) 14964). Lectures Mon at 2, Tues at 4, in B1. Office hours: displayed outside my office. (see notices and timetable outside my room). AIMS AND OBJECTIVES: Aims: to teach the introductory theory of functions of a complex variable; to teach the computational techniques of complex analysis, in particular residue calculus, with a view to potential applications in subsequent modules. Objectives: a successful student will: 1. be able to identify analytic functions and singularities; 2. be able to prove simple propositions concerning functions of a complex variable, for example using the Cauchy-Riemann equations; 3. be able to evaluate certain classes of integrals; 4. be able to compute Taylor and Laurent series expansions. SUMMARY: in this module we concentrate on functions which can be regarded as functions of a complex variable, and are differentiable with respect to that complex variable. These "good" functions include exp, sine, cosine etc. (but log will be a bit tricky). These are important in applied maths, and they turn out to satisfy some very useful and quite surprising and interesting formulas. For example, one technique we learn in this module is how to calculate integrals like + ∞ cos x ∫ − ∞ x 2 + 1 dx WITHOUT actually integrating. PROBLEM CLASSES will be fortnightly, on Tuesdays at 11.00 and 12.00. You must be available for at least one of these times. Please see handouts for dates and further information. COURSEWORK: Dates for handing in for G12CAN will be announced in the first handout (all will be Tuesdays). The problems will be made available at least one week before the work is due. Homework does not count towards the assessment, but its completion is strongly advised, and the work will emphasize the computational techniques which are essential to passing the module. Failure to hand in homework, poor marks, and non-attendance at problem classes will be reported to tutors. ASSESSMENT: One 2-hour written exam. Section A is compulsory and is worth half the total marks. From Section B you must choose two out of three longer questions. For your revision, you may find it advantageous to look at old G12CAN papers, although there have been minor variations in content over the years. The assessment will mainly be based on using the facts and theorems of the module to solve problems of a computational nature, or to derive facts about functions. You will not be expected to memorize the proofs of the theorems in the notes.
¢ £ ¢ ¡
-2Proofs of some theorems will just be sketched in the lectures, with the details provided on handouts in case you wish to see them. You will not be required to reproduce these proofs in the examination. 1.1 Basic Facts on Complex Numbers from G1ALIM All this section was covered in G1ALIM. Suppose we have two complex numbers z = x + yi and w = u + i (where x, y, u, are all real). Then x = Re(z ), y = Im( z ), (x + yi) + (u + i) = (x + u) + ( y + )i, (x + yi) − (u + i) = (x − u) + ( y − )i, (x + yi)(u + i) = xu − y + (x + yu) i and, if x + yi = 0 + 0i , / u+ i (u + i)(x − yi) (u + i)(x − yi) = = . 2 2 x + yi x +y (x + yi)(x − yi) , which contains
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With these rules, we’ve made a field called
, as x = x + 0i .
The Argand diagram, or complex plane Think of the complex number z = x + yi , with x = Re( z ), y = Im( z ) both real, as interchangeable with the point (x, y) in the two dimensional plane. A real number x corresponds to (x, 0) and the x axis becomes the REAL axis, while numbers iy , with y real (often called purely imaginary) correspond to points (0, y), and the y axis becomes the IMAGINARY axis. The complex conjugate The complex conjugate of the complex number z is the complex number z = Re( z ) − i Im( z ). Some write z* instead. E.g. 2 + 3i = 2 − 3i . In fact, z is the reflection of z across the real axis. The conjugate has the following easily veri fied properties: (z ) = z , z + w = z + w, z w = z w, z + z = 2Re( z ), z − z = 2iIm( z ). Modulus of a complex number
( ( ( ( ( ( ( ( ( ( ( £¢£¢§¢£¢£§¢( % $ ## " ! ! ! © Q
The modulus or absolute value of z is the non-negative real number z = is the distance from 0 to the point z in the complex plane. Note that
0 ' & )
√Re(z)2 + Im( z)2. This
2
1
zz = (Re( z ) + iIm( z ))(Re( z ) − iIm( z )) = Re(z )2 + Im( z )2 = z
A @ 9 X 8 7 4 6 54
Warnings (i) The rules z = ± z , z 2 = z 2 are only true if z is real; (ii) The statement z < w only makes sense if z and w are both real: you can’t compare complex numbers this way. Triangle Inequality For all z ,w ∈ , we have z + w z + w and z − w z − w . Note that 0, z , w,z + w form the vertices of a parallelogram. The second inequality follows from z w + z −w .
g f e d c b a ` Y W V U T S R P I H G F
E
CDC
B
so that a useful formula is z =
3 2
√zz. Also (i) 1 / z = z z
−2
if z = 0 (ii) z w = z /
w .
-3Note also that 0, w, z − w, z form the vertices of a parallelogram and hence z − w is the distance from z to w . Polar and exponential form Associate the complex number z with the point (Re( z ) , Im( z )) in 2. If z ≠ 0, then Re(z ) and Im( z ) aren’t both zero, and r = z = 0. Let θ be the angle between / the positive real axis and the line from 0 to z , measured counter-clockwise in radians. Then x = Re(z ) = r cos θ , y = Im( z ) = r sin θ . Writing
i h
z = r cos θ + ir sin θ ,
we have the POLAR form of z . The number θ is called an ARGUMENT of z and we write θ = arg z . Note that (1) arg 0 does not exist. (2) If θ is one argument of z , then so is θ + k 2π for any integer k . (3) From the Argand diagram, we see that arg z ± π is an arg of − z . We can always choose a value of arg z lying in (− π, π ] and we call this the PRINCIPAL ARGUMENT Arg z . Note that if z is on the negative real axis then Arg z = π, but Arg z → − π as z approaches the negative real axis from below (from the lower half-plane). To compute Arg z using a calculator: suppose z = x + iy = 0, with x, y real. If x > 0 then / −1 θ = Arg z = tan (y / x) = arctan( y / x) but this gives the WRONG answer if x < 0. The reason is that calculators always give tan − 1 between − π / 2 and π / 2. Thus if x < 0 then tan − 1( y /x) = tan − 1( −y / (− x)) gives Arg ( − z ) = Arg z ± π. If x = 0 and y > 0 then Arg z = π / 2, while if x = 0 and y < 0 then Arg z = − π / 2. Definition For t real, we define e it = cos t + i sin t . Using the trig. formulas cos(s + t) = cos s cos t − sin s sin t, sin(s + t) = sin s cos t + cos s sin t, we get, for s,t real, e ise it = cos s cos t − sin s sin t + i(cos s sin t + sin s cos t) = e i(s + t). Thus e − ite it = e i 0 = 1. Also, (e it) = e − it and, if z ,w are non-zero complex numbers, we have
rrrr
z w = z e i arg z w e i arg w = z w e i (arg z + arg w)
and z = z e − iarg z, 1 / z = z − 1e − iarg z. We get: (a) arg z + arg w is an argument of z w. (b) − arg z is an argument of 1 / z and of z . Warning: it is not always true that Arg z + Arg w = Arg z w. Try z = w = − 1 + i . De Moivre’s theorem For t real, we have e 2it = e ite it = (e it) 2 and e − it = 1 / (e it). Repeating this argument we get (e it)n = e int for all real t and integer n (de Moivre’s theorem). For example, for real t , we have cos 2t = Re(e 2it) = Re((e it) 2) = Re(cos2t − 2icos tsin t − sin 2t) = 2cos2t − 1.
y x w v u t s
q
p
-4Roots of unity Let n be a positive integer. Find all solutions z of z n = 1. Solution: clearly z = 0 so write z = re it with r = z and t an argument of z . Then / n n int 1 = z = r e . So 1 = z n = r n and r = 1, while e int = cos nt + i sin nt = 1. Thus nt = k 2π for some integer k , and z = e it = e k 2πi/ n. However, e is = e is + j 2πi for any integer j , so e k 2πi / n = e k ′ 2πi / n if k − k ′ is an integer multiple of n . So we just get the n roots ζ k = e k2πi /n, k = 0, 1,...., n − 1. One of them (k = 0) is 1, and they are equally spaced around the circle of centre 0 and radius 1, at an angle 2π / n apart. The ζ k are called the n ’th roots of unity. Solving some simple equations To solve z n = w , where n is a positive integer and w is a non-zero complex number, we first write w = w e i Arg w. Now z 0 = w 1/ ne (i / n)Arg w, in which w 1/ n denotes the positive n ’th root of w , gives ( z 0 ) n = w. This z0 is called the principal root. Now if z is any root of z n = w , then ( z / z 0 ) n = w/ w = 1, so z / z 0 is an n ’th root of unity. So the n roots of z n = w are z k = w 1/ ne (i / n)Arg w + k 2πi/ n, k = 0, 1, . . . , n − 1. For example, to solve z 4 = − 1 − i = w, we write w = √2e − 3πi/ 4 and z 0 = 21/ 8e − 3πi /16. The other roots are z 1 = 21 / 8e − 3πi/ 16 + πi/ 2 = 2 1/ 8e 5πi/ 16 and z 2 = 2 1/ 8e − 3πi/ 16 + πi = 21 / 8e 13πi / 16 and z 3 = 21 / 8e − 3πi/ 16 + 3πi /2 = 21 / 8e − 3πi/ 16 − πi / 2 = 21 / 8e − 11πi / 16. Quadratics: we solve these by completing the square in the usual way. For example, to solve z 2 + (2 + 2i) z + 6i = 0 we write this as (z + 1 + i)2 − (1 + i)2 + 6i = 0 giving (z + 1 + i)2 = − 4i = 4e − i π / 2 and the solutions are z + 1 + i = 2e − i π / 4 and z + 1 + i = 2e − i π /4 + i π = 2e 3i π / 4. In general, az 2 + bz + c = 0 (with a = 0) solves to give 4a 2z 2 + 4abz + 4ac = 0 and so (2az + b) 2 = / 2 2 1 /2 b − 4ac and so z = ( − b + (b − 4ac) ) / 2a with, in general, two values for the square root. For example, to solve z 4 − 2z 2 + 2 = 0 we write u = z 2 to get (u − 1)2 + 1 = 0 and so u = 1 ± i . Now z 2 = 1 + i = √2e i π / 4 has principal root z1 = 21 / 4e i π / 8 and second root z 2 = z 1 e i π = − z1 = 21 /4e i 9π /8 = 21 / 4e − i 7π / 8, in which 21 / 4 means the positive fourth root of 2. Two more solutions come from solving z 2 = 1 − i = √2e − i π / 4 and these are z 3 = 2 1/ 4e − i π / 8 and z 4 = 21 / 4e i 7π / 8.
e d
-5An example Consider the straight line through the origin which makes an angle α ,0 α π / 2, with the positive x -axis. Find a formula which sends each z = x + iy to its reflection across this line. If we do this first using the line with angle α , and then using the line with angle β ( 0 < β < π / 2 ), what is the net effect? 1.2 Introduction to complex integrals Suppose first of all that [a,b] is a closed interval in and that g :[a,b] → is continuous (this means simply that u = Re(g) and = Im(g) are both continuous). We can just define
∫
Example Determine
b
g(t) dt =
a
∫
2
e 2itdt . 0 and t ∈ ,
g f
0
Note that every complex number z can be written in the form z = re it with r = z and so z = z e − it. Thus we have, for some real s,
k j i h
∫
b
g(t) dt = e is
a
and this is, by real analysis,
q o m
Re( e isg(t) )
dt
e isg(t)
dt =
a
a
a
Example: for n ∈
set In =
∫
2
e it (t + in) − 1dt . Show that I n → 0 as n → + ∞.
3
1
1.3 Paths and contours Suppose that f1 , f2 are continuous real-valued functions on a closed interval [a,b]. As the "time" t increases from a to b , the point γ (t) = f1(t) + if2(t) traces out a curve ( or path, we make no distinction between these words in this module ) in . A path in is then just a continuous function γ from a closed interval [a,b] to , in which we agree that γ will be called continuous iff its real and imaginary parts are continuous. Paths are not always as you might expect. There is a path γ :[0,2] → such that γ passes through every point in the rectangle w = u + i , u, ∈ [0,1]. (You can find this on p.224 of Math. Analysis by T. Apostol). There also exist paths which never have a tangent although (it’s possible to prove that) you can’t draw one. Because of this awkward fact, we define a special type of path with good properties: A smooth contour is a path γ :[a, b] → such that the derivative γ ′ exists and is continuous and
¤ ¤
p
n
l
∫
b
¤
∫
b
Re(g(t)) dt + i
a
∫
b
Im(g(t)) dt.
a
∫
b
g(t) dt =
a
∫
b
b
e isg(t) dt =
a
∫
b
Re( e isg(t) ) dt
a
∫
∫
b
g(t)
dt.
-6never 0 on [a, b]. Notice that if we write Re(γ ) = σ , Im(γ ) = τ then (σ ′ (t) , τ ′ (t)) is the tangent vector to the curve, and we are assuming that this varies continuously and is never the zero vector. For a < t < b let s(t) be the length of the part of the contour γ between "time" a and "time" t . Then if δ t is small and positive, s(t + δ t) − s(t) is approximately equal to γ (t + δ t) − γ (t) and so
δt → 0 +
{ z x£x£x£x¢x£x¢x§x£x¢x£x¢xyx
lim
a
Examples (i) A circle of centre a and radius r described once counter-clockwise. The formula is z = a + re it, 0 t 2π. (ii) The straight line segment from a to b . This is given by z = a + t(b − a), 0 t 1. More on arc length (optional!) Let γ :[a, b] → , γ (t) = f (t) + ig(t), with f, g real and continuous, be a path (not necessarily a smooth contour). The arc length of γ , if it exists, can be defined as follows. Let a = t0 < t1 < t2 < ...... < tn = b . Then P = {t0 , ......, tn } is a partition of [a, b] with vertices tk (the notation and some ideas here have close analogues in Riemann integration), and
L(P ) = ∑ γ (tk ) − γ (tk − 1 )
k =1
is the length of the polygonal path through the n + 1 points γ (tk ), k = 0, 1, . . . , n . If we form P ′ by adding to P an extra point d , with tj − 1 < d < tj , then the triangle inequality gives
L(P ′) − L(P) = γ (tj ) − γ (d) + γ (d) − γ (tj − 1 ) − γ (tj ) − γ (tj − 1 )
So as we add extra points, L(P ) can only increase, and if the arc length S of γ exists in some sense then it is reasonable to expect that L(P ) will be close to S if P is "fine" enough (i.e. if all the tk − tk − 1 are small enough). With this in mind, we define the length S of γ to be S = Λ (γ , a, b) = l.u.b. L(P ), with the supremum (l.u.b. i.e. least upper bound) taken over all partitions P of [a, b]. If the L(P) are bounded above, then S is the least real number which is L(P) for every P, and γ is called rectifiable. If the set of L(P ) is not bounded above then S = ∞ and γ is non-rectifiable. Suppose that a < c < b . Then every partition of [a, b] which includes c as a vertex can be written as the union of a partition of [a, c] and a partition of [c, b]. It follows easily that
Λ (γ , a, b) = Λ (γ , a, c) + Λ (γ , c, b).
The following theorem shows that, for a smooth contour, the arc length defined this way has the
n
0.
~
}
Hence the length of the whole contour γ is
|
∫
b
γ ′ (t) dt , and is sometimes denoted by γ .
w
ds = dt
γ (t + δ t) − γ (t) δt
= γ ′ (t) .
s
r
v
tut
-7
a
Theorem Let γ :[a, b] →
, γ = f + ig , with f,g real, be a smooth contour. Then S as defined above satisfies
S = Λ (γ , a, b) =
a
Proof: Let
¤
S( ) = Λ (γ , a, ), a
¤
If we can show that S ′ ( ) = γ ′ ( )
¤ ¤ ¤ ¤
for a <
2 2
< b then (1) follows by integration. So let
a < < b and let c = γ ′ ( ) = √f ′ ( ) + g ′ ( ) . We know that c = 0 (definition of smooth con/ tour). Let 0 < δ < c , and choose ε > 0, so small that − ε < p < ε and − ε < q < ε imply that c − δ < √( f ′ ( ) + p) 2 + (g ′ ( ) + q) 2 < c + δ . Since γ ′ = f ′ + ig ′ is continuous at , we can choose ρ > 0 such that f ′(s) − f ′( ) < ε , g ′(s*) − g ′ ( ) < ε ,
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Let 0 < h < ρ , and let P = {t0 , t1 , . . . , tn } be any partition of [ , + h]. Then L(P ) = ∑ γ (tk ) − γ (tk − 1 ) = ∑ √( f (tk ) − f(tk − 1 ))2 + (g(tk ) − g(tk − 1 )) 2
k =1 k =1
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n
n
and the mean value theorem of real analysis gives us s k and s k * in (tk − 1 , tk ) such that L(P ) = ∑ (tk − tk − 1 ) √f ′ (sk )2 + g ′ (sk *) 2.
k =1 n
§ § § § § § § § § § § § § £££¢£¢§£¢£¢§¨§
Hence, by (4), we have (tn − t0 )(c − δ ) = ∑ (tk − tk − 1 )(c − δ ) < L(P ) < ∑ (tk − tk − 1 )(c + δ ) = (tn − t0 )(c + δ ).
k =1 k =1 n n
Since P is an arbitrary partition of [ , + h] we get
¤ ¤ ¤ ¤
h(c − δ ) and so, provided 0 < h < ρ ,
Λ (γ , , + h) = S( + h) − S( )
©¢©£©¢©£©£©£©¢©£©¢©§ ©©
c−δ
S( + h) − S( ) h
¤ ¤
c+δ.
¤
¤
¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢£¢§£¢£¢§¢£¢
c−δ <
√f ′ (s)2 + g ′ (s*)2 < c + δ .
¤
¤
¤
¤
¤
for s −
< ρ , s* −
< ρ . So, for s −
< ρ , s* −
¤
¤
same value as the integral
∫
b
γ ′ (t) dt which we derived earlier.
∫
b
γ ′(t) dt.
(1)
b.
(2)
(3)
< ρ , (2) and (3) give (4)
h(c + δ )
-8Since δ may be chosen arbitrarily small we get S ′( ) = c = γ ′ ( ) . 1.4 Introduction to contour integrals Suppose that γ :[a, b] → is a smooth contour. If f is a function such that f (γ (t)) is continuous on [a, b] we set
« ¤ ª ¤
∫
γ
f ( z ) dz =
∫
b
f(γ (t)) γ ′ (t) dt.
a
1.5 a very important example! Let a ∈ , let m ∈ and r > 0, and set γ (t) = a + re it , 0 t 2m π. As t increases from 0 to 2m π, the point γ (t) describes the circle z − a = r counter-clockwise m times. Now let n ∈ . We have
∫
γ
( z − a)n dz =
∫
2m π
r ne int ire it dt =
0
If n ≠ − 1 this is 0, by periodicity of cos ((n + 1)t) and sin ((n +1) t). If n = − 1 then we get 2m πi . 1.6 Properties of contour integrals (a) If γ :[a,b] → is a smooth contour and λ is given by λ (t) = γ (b + a − t) (so that λ is like γ "backwards") then
∫
λ
f ( z ) dz =
∫
b
f(γ (b + a − t)) ( − γ ′ (b + a − t)) dt = −
a
(b) A smooth contour is called SIMPLE if it never passes through the same point twice (i.e. it is a one-one function). Suppose that λ and γ are simple, smooth contours which describe the same set of points in the same direction. Suppose λ is defined on [a,b] and γ on [c,d]. It is easy to see that there is a strictly increasing function φ :[a,b] → [c,d] such that λ (t) = γ (φ (t)) for a t b . Further, it is quite easy to prove that φ (t) has continuous non-zero derivative on [a,b] and we can write
∫
λ
f ( z ) dz =
∫
b
f(λ (t)) λ ′(t) dt =
a
=
∫
d
f(γ (s)) γ ′(s) ds =
c
Thus the contour integral is "independent of parametrization". Here’s the proof that φ ′(t) exists (optional!). For t and t0 in (a, b) with t = t0 write /
¬
∫
2m π
i r n + 1 e (n + 1) it dt.
0
∫
γ
f ( z ) dz .
∫
b
f (γ (φ (t))) γ ′ (φ (t)) φ ′ (t) dt =
a
∫
γ
f ( z ) dz .
-9λ (t) − λ (t0 ) γ (φ (t)) − γ (φ (t0 )) φ (t) − φ (t0 ) = . t − t0 φ (t) − φ (t0 ) t − t0
±¢±£±¢±£±£±£±¢±²± °§°¢°£°¢°£°§°¢°£°¢°£°£°£ °°
Note that there’s no danger of zero denominators here as φ is strictly increasing so that φ (t) = φ (t0 ). Letting t tend to t0 we have γ (φ (t)) → γ (φ (t0 )) and so φ (t) → φ ..."
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