Table Of ContentA solutions manual for Complex Analysis
In the Spirit of Lipman Bers
by Rub´ı E. Rodr´ıguez, Irwin Kra and Jane Gilman
Vamsi P. Pingali
Preface
About the manual
This manual is intended to accompany the second edition of Complex Analysis in the spirit
of Lipman Bers by R. Rodriguez, I. Kra and J. Gilman. It consists of solutions to most
of the problems. Some problems involving filling in the details of a proof in the text and
those involving the proof of a well known theorem that can be found in other texts have
been omitted. Also, in some cases, wherein the procedure is clear and mechanical, an outline
of the solution is given (as opposed to a complete one). In order to gain most out of the
problems, readers are strongly encouraged to try solving the problems themselves before
referring to the manual.
Acknowledgements
I thank Irwin Kra for delegating this task to me and for helping me out with it. I also
thank R. Rodriguez and J. Gilman for their suggestions regarding the problems and other-
wise. MyappreciationalsoextendstomyadviserLeonTakhtajan, myfriendsandmyfamily.
Contact
The author would be most grateful if mistakes (whether typographical or mathematical) are
pointed out or if suggestions are given. To do so, please feel free to email the author at
[email protected]. The author’s current webpage (where errors shall be posted) is
www.math.sunysb.edu/ vpingli
∼
Vamsi Pritham Pingali,
Department of Mathematics,
Stony Brook university.
1
Chapter 2
1. (a) If we fix m, then z z < ǫ n > m 1+1. Hence z converges to z as n .
| n− m| ∀ − ǫ n m → ∞
But this is true for all m. Thus, z is a constant sequence (and can be any constant
m
sequence. We don’t have enough information to discern the value of the constant).
(b) Let C be such that w C. Thus, if n > N, z < ǫ and hence w z < ǫ.
| n| ≤ | n| C | n n|
Thus proved.
2. (a) Noting that b 2 = b¯b,
| |
c¯z 1 2 z c 2 = cz 2 +1 c¯z z¯c z 2 c 2 +c¯z +z¯c
| − | −| − | | | − − −| | −| |
= cz 2 +1 z 2 c 2
| | −| | −| |
= (1 z 2)(1 c 2)
−| | −| |
(b) If z c < c¯z 1 , then, by (a), (1 z 2)(1 c 2) > 0, and hence, z < 1 (because
| − | | − | −| | −| | | |
c < 1). Conversely, if z < 1, then z c < c¯z 1 . The remaining parts of the
| | | | | − | | − |
problem are similar.
3. There is a t [0,1] so that b = ta+(1 t)c. In order to show that a complex number
∈ −
c is real, all one needs to do is to prove that c = c¯. Indeed,
¯
a b a¯ b (a ta (1 t)c)(c¯ ta¯ (1 t)c¯) (c ta (1 t)c)(a¯ ta¯ (1 t)c¯)
− − = − − − − − − − − − − − − −
c b − c¯ ¯b c b 2
− − | − |
= 0
4. (a) Every z that lies in the triangle lies on a line joining 1 to the side containing z
1
and z2. Let −(α+1) be a point on the line. Note that δ = z|α|1 ≥ 1. Hence,
| − |
1 z α
| − | = | |
1 z δ α+δ
−| | −| |
tz +(1 t)z 1
1 2
= | − − |
δ δ 1+tz +(1 t)z
1 2
−| − − |
where 0 t 1. We see that the above expression is decreasing as a function of delta
≤ ≤
(quite easily by computing the derivative). Hence, it is maximum when δ = 1; i.e.,
when z is on the line joining z and z . Maximising the resulting expression gives us
1 2
2
the K. This expression has a maximum because it is a continuous function of t (on a
compact set [0,1]).
(b) The maximum is attained when z = 1+i. This is because 1 z is largest for this
2 | − |
z, and z is also largest for this z. Hence, K = 1 .
| | √2 1
−
5. f(z) = z3 = (x+ıy)3 = x3 3y2x+ı( y3 +3x2y). Hence, u = 3x2 3y2 = v and
x y
− − −
u = 6xy = v .
y x
− −
6. Let z = t(a+ıb) be a straight line connecting a+ıb to 0 (where a, b are real).
f(z) f(0) t4ab2(a+ıb)
lim − = lim
z t 0 t(a+ıb)(t2a2 +t4b4)
→
tab2
= lim
t 0 a2 +t2b4
→
= 0
Plugging in x = y2 in f,
f(z) f(0) y4(y2 +ıy)
lim − = lim
z y 0 (y2 +ıy)2y4
→
1
=
2
Hence f′(0) does not exist (the limit is different along different paths).
7. By the chain rule, (note that x2 +y2 = r2 and y = tan(θ) when x = 0)
x 6
∂r ∂θ
u = u +u
x r θ
∂x ∂x
sin(θ)u
θ
= cos(θ)u
r
− r
v = v r +v θ
y r y θ y
cos(θ)v
θ
= v sin(θ)+
r
r
cos(θ)u
θ
u = u sin(θ)+
y r
r
sin(θ)v
θ
v = cos(θ)v
x r
− r
By imposing the usual CR equations (i.e. u = v and u = v ) and solving for u
x y y x r
−
and u in terms of v and v , we get the result for x = 0. One may prove this for x = 0
θ r θ
6
by considering cot(θ) = x and going through the same argument.
y
8. Since f is C1, being holomorphic is equivalent to the Cauchy-Riemann equations.
Hence, f = ( ∂ +ı ∂ )(u+ıv) = (u v )+ı(u +v ) which equals 0 if and only if
z¯ ∂x ∂y x − y y x
the CR equations are satisfied. Since (in this case) f′ = u +ıv , f′ = f .
x x z
3
9. By the previous problem, since f is C1, f is holomorphic if and only if f = 0, which
z¯
is the same as R eıΦ + ıReıΦΦ = 0 (notice that the usual product rule carries over
z¯ z¯
trivially to ∂ and ∂ ). The last statement is equivalent to R +ıRΦ = 0.
∂z¯ ∂z z¯ z¯
10. Let f = u+ıv. We shall prove this for (g f) using the usual Chain rule for derivatives
z
◦
(the other case is similar) :
(g f) = (g f) ı(g f)
z x y
◦ ◦ − ◦
= g u +g v ı(g u +g v )
x x y x x y y y
−
1
¯ ¯ ¯ ¯
= (g (f +f ) ıg (f f ) ı(g (f +f ) ıg (f f )))
x x x y x x x y y y y y
2 − − − − −
¯
= g f +g f
w z w¯ z
11. p is an entire function if and only if p = 0. Thus, P (z)jz¯j 1 = 0 for all z¯.
z¯ j −
j 1
X≥
Comparing coefficients, we see that the condition is equivalent to P = 0 for all j 1.
j
≥
N
If indeed p is an entire function, then p(z) = P (z) = c zk. This implies that
0 k
k 0
X≥
N µ=k
k!
p(z) = c xkık µyk µ. Hence, interchanging the order of summation
k − −
(k µ)!µ!
k 0 µ=0 −
X≥ X
we see that a = ıνc (µ+ν)! when 0 ν N µ and 0 µ N (otherwise
µν µ+ν µ!ν! ≤ ≤ − ≤ ≤
a = 0).
µν
12. f is anti-holomorphic if f is holomorphic. Hence, if f = u+ıv, (by replacing v by v
−
in the usual CR equations), u = v and u = v . In polar coordinates, if f = U+ıV,
x y y x
−
the equations are rU = V and rV = U .
r θ r θ
−
If f is C1, the equation for f¯to be holomorphic is f¯ = 0. But it is easy to see from the
z¯
¯
definitionsthat f = (f ), and therefore in this case the condition f is anti-holomorphic
z¯ z
is equivalent to f = 0.
z
¯
13. g (z) = (f(z¯)) = f(z¯) = 0; the last statement follows by applying the Chain rule.
z¯ z¯ z
Similarly, h = f(z¯) = 0 (by the same argument as before). Hence, g is holomorphic
z z
and h is anti-holomorphic. If h is also holomorphic, then h = 0. Thus, h = h = 0.
z¯ x y
Hence, h = f(z¯) is a constant which implies f is too.
14. If f is holomorphic and anti-holomorphic, then f = f = 0 and hence f = f = 0.
z z¯ x y
Since D is connected, f is a constant (if it weren’t connected, f could be different
constants on the connected components).
15. (a) No. If f = u + ıv is holomorphic and C2, then u + u = v v = 0. But,
xx yy xy yx
−
clearly u = ex is C2 and does not satisfy the condition and hence can’t be the real part
of a holomorphic function.
4
(b) Yes. The functions f = ex(xcos(y) ysin(y))+ıex(xsin(y)+ycos(y))+ıb, with b
−
in R, are holomorphic (the real and imaginary parts are smooth functions that satisfy
the Cauchy-Riemann equations) and their real part is the given function. In order to
derive the formula for these functions, we shall solve the Cauchy-Riemann equations:
u = ex(xcos(y) ysin(y)+cos(y))
x
−
v = u
y x
v = u dy +g(x)
x
Z
= ex(xsin(y)+sin(y))+ exysin(y)dy +g(x)
Z
= ex(xsin(y)+sin(y)) ex( ycos(y)+sin(y))+g(x)
− −
= ex(xsin(y)+ycos(y))+g(x)
One may see (quite easily) that the second Cauchy-Riemann equation (i.e. u = v )
y x
−
implies that g (x) = 0; i.e., g(x) is a real constant.
′
16. We shall follow the hints in the text :
(a) Firstly, note that p(z) zn a zn 1+...+a . Now, zk zn 0 k n
n 1 − 0
when z 1. Thus, |a |z≥n |1|+|.−..|+−a zn 1 ( a || +...|+ |a≤)| =|C∀ zn≤1 .≤So,
n 1 − 0 − n 1 0 −
p(z) | |C≥zn 1 > M|wh−en z > R = ma|x≤(2C| ,1,(|M|)1−/(n| 1)). | | | |
| | ≥ | − | | | C −
(b) Since S = (z; z R) is compact, p(z) (being continuous) attains a minimum
| | ≤ | |
over it. Since p(z) M for z C S, the minimum of p(z) over C is attained at
| | ≥ ∈ − | |
some point z .
0
(c) Changing variables we define g(z) = p(z + z ), clearly g(0) = 0 if and only if
0
p(z ) = 0.
0
(d) Write g(z) = α + βzmh(z) (where h = h˜ + 1 is a polynomial such that h˜(0) = 0
and β = 0). Note that g(0) = α is the minimum of g . But, if α = 0, then defining
6 | | | | 6
γm = α, g(γz) = α 1 zmh(γz) < α for some small, positive, real z. Indeed,
−β | | | || − | | |
1 zm(h˜(γz) + 1) 1 zm + zm h˜(γz) < 1 for a small positive, real z, because
|˜ − | ≤ − | |
h(γz) is small. This is a contradiction and hence g(0) = 0.
| |
17. We shall prove this by induction: The base case is trivial. If the statement is true for
k = n, then, ifp(z)isadegreen+1polynomial, bythefundamentaltheoremofalgebra,
it has a root and hence p(z) = (z z )h(z) where h(z) is a degree n polynomial. By the
0
−
induction hypothesis, h(z) = (z z )...(z z ) and hence p(z) = (z z )(z z )....
1 n 0 1
− − − −
This implies that the number of roots (counted with multiplicities) is indeed n.
Given a real polynomial p(x) = a xn, we see that p(z¯) = p(z). Hence, the non-
n
real roots (if any) occur in conjugate pairs. Since every nonconstant degree d real
P
polynomial can be factored into complex linear factors p(x) = a (x z )(x z¯ )... =
d 1 1
− −
a (x2 (z +z¯ )x+ z 2)..., it can be factored into products of real linear or quadratic
d 1 1 1
− | |
factors.
5
18. Once again we shall follow the hints in the text :
(a) Consider θ F R. The set 1,θ,θ2,...θn 1,θn must be linearly dependent over
−
∈ − { }
R (because the dimension of F is assumed to be n). Hence there exists a non-constant
real polynomial p(x) = xk +a xk 1 +..., so that p(θ) = 0.
k 1 −
−
(b) Every real polynomial can be factored into a product of quadratics and linear
factors (by the previous problem). Hence, there exist real numbers β and γ so that
θ2 2βθ+γ = 0.
−
(c) By (b) (θ β)2 = β2 γ. If the right hand side is positive, then we could’ve
− −
factored the quadratic into real linear factors, but θ is not real. Hence, β2 γ = δ2
− −
for some positive real number δ. Therefore, σ = θ β satisfies σ2 = 1.
−
δ −
(d) Clearly, the field G in the text is isomorphic to C by sending σ ı. Since very
→
element of F satisfies a complex (in fact real) polynomial by (b), it follows from the
fundamental theorem of algebra that F = G.
19. (a) Every automorphism σ takes 1 to 1. Hence, it is the identity on Q. If x > 0, then
σ(x) = (σ(√x))2 > 0. Thus, σ preserves the order relation on the reals. If we assume
that σ(x) < x for some real x (without loss of generality), then σ(x) < a x for some
≤
a Q. But, if a x, then σ(a) = a σ(x) (it preserves the order). A contradiction.
∈ ≤ ≤
Hence, σ is the identity.
(b) An automorphism σ fixing the reals takes a root of the polynomial x2 +1 = 0 to
another root. Hence, it takes ı to ı or to ı. This means it maps bı to σ(b)σ(ı) = bı
−
or to bı for all b in R. Thus, σ(a+ıb) = a+ıb or a ıb for all a in R. The former
− −
corresponds to the identity map and the latter to conjugation.
(c) Note that σ(1) = 1 and hence σ restricted to the rationals is the identity. For
every real r, there exists a sequence of rationals a r. Since σ is continuous,
n
→
a = σ(a ) σ(r) and hence σ maps reals to reals. By (a), σ restricted to the reals
n n
→
is the identity. By (b), we are done.
20. Yes. By general topology (see Munkres for instance), every connected, locally path
connected space is path connected. Every open set in C is obviously locally path
connected (there is an open disc around every point). Hence connectedness and path-
connectedness are the same for open sets in C.
6
Chapter 3
1. We shall use the ratio test: R = lim an where R is the radius of convergence.
n→∞ an+1
We see (quite easily) that the radii of convergence are , 1, 1 and 0 (i.e the last series
∞
does not converge anywhere except at z = 0) respectively.
(a) The series zn is absolutely convergent even when z = 1 (because 1 is).
On the boundanr2y, the series zn converges when z =|1|and diverges whennz2 = 1
P n 6 P
(by the integral test). One may prove this by observing that this series equals
P
ln(1 z). To prove the statement without using this observation, the reader is
−
referred to Rudin’s book “Principles of Mathematical Analysis” (the idea is to
use summation by parts to prove the assertion).
(b) The series ez = zn does not converge at infinity because, as z via the
n! → ∞
negative reals ez 0, and along the positive reals it goes to .
P→ ∞
2. The given series is 1 as is easily seen by multiplying it with z2+z 1. Hence, it
z2+z 1 −
converges when z is le−ss than the roots of z2 + z 1 (by theorem 3.55). Thus, the
| | −
radius equals √5 1.
−
4
3. Since a M, lim ( a )1/n 1 and hence ρ 1 (by Hadamard’s theorem).
n n n
| | ≤ →∞ | | ≤ ≥
4. (a) If one of the sequences is unbounded from above (and the other does not converge
to 0), then the right hand side is and the result is trivial. If both of them are
∞
bounded from above, the limsups are finite; sup a b sup a sup b . Hence,
n≥p n n ≤ n≥p n n≥p n
inf sup a b inf (sup a sup b ) = lim (sup a sup b ) where the last
p n≥p n n ≤ p n≥p n n≥p n p n≥p n n≥p n
equality holds because (sup a sup b ) is decreasing. So, inf sup a b
lim sup a lim sup b na≥npdnthe nr≥espulnt follows. If a = 2 + (p 1)nn≥apndnbn ≤=
5+p( 1)nn≥+p1,nthe lepft hann≥dp snide is 12 whereas the right handn side is 18−. n
−
(b) If b is not bounded from above, then sup a b = for every p (a gets close to
n n≥p n n ∞ n
thelimitandb hasasubsequencegoingto ). Hencetheresultholdsinthiscase. Ifb
n n
∞
isboundedfromabove,thenthereexistsasubsequenceb goingtosup b foragiven
nk n p n
≥
p. Hence, lim (a b ) = lima sup b i.e. lim (a b ) lima limsupb .
k→∞ nk nk n n≥p n k→∞ nk nk ≥ n n
By problem 18, the limsup is the supremum of all subsequential limits. Hence the
resut follows.
5. If lim an+1 = L, then (L ǫ)a a (L+ǫ)a n N. Hence, (L ǫ)n N+1a
an − n ≤ n+1 ≤ n∀ ≥ − − N ≤
a (L+ǫ)n N+1a . Using the geometric series, this implies that a converges
n+1 − N n
≤
P
7
when L+ǫ < 1 for an ǫ > 0 and diverges when L ǫ 1 for an ǫ > 0 i.e. the series
− ≥
converges when L < 1 and diverges when L > 1.
6. If a zn and b zn converge (which they do when z < ρ), then, (a + b )zn
n n n n
| |
converges to a zn + b zn:
n n
P P P
n=NP P
∞ ∞ ∞ ∞
(a +b )zn ( a zn + b zn) = a zn + b zn
n n n n n n
| − | | |
n=0 n=0 n=0 n=N n=N
X X X X X
∞ ∞
a zn + b zn
n n
≤ | | | |
n=N n=N
X X
ǫ ǫ
+
≤ 2 2
for sufficiently large N. Also, a b zn = c zn converges to ( a zn)( b zn)
i i n i n n n
−
by lemma 3.17. Hence, the radii of convergence of these two series are ρ.
PP P P≥ P
7. Changing variables to w = 1 and apply corollary 3.60 to f(1) to conclude that,
z w
f(1) = wng. Hence, there exists a ball B centred at z = 0 such that outside the ball,
w
f(z) = 1 g(z) where g is a holomorphic function on C B and is a unit.
zn −
8. We shall prove that e zez = 1. By theorem 3.18, we are allowed to multiply the power
−
series (like polynomials). Hence, e−zez = ( (−1n)!nzn)( zmm!) = k nn==k0 n(−!(1k)nnz)k! =
−
zk(1 1)k = 1.
k k! − P P P P
9. (Pa) Certainly, a branch of f(z) = √z is an algebraic function (which is clearly neither
a polynomial nor a rational function). Indeed, (f(z))2 z = 0.
−
(b) Assume otherwise. Then, there exist polynomials with rational coefficients so that
p (z)enz = 0. Substitute z = 1 in this equation to deduce that e is algebraic, a
n
contradiction.
P
10. No. Log(eı2π) = ln e2πı +ıArg(e2πı) = 0 = 2πı.
| | 6
11. If exp(exp(z)) = 1, then exp(z) = 2nπı where n is a non-zero integer. Hence, z =
ln(2 n π) + ı(( 1)sgn(n)π + 2kπ) where k Z and sgn(n) = 0 if n is positive and 1
| | − 2 ∈
otherwise.
12. (a) Let the period of cos(z) be p: eı(z+p) + e ı(z+p) = eız + e ız i.e. eız(eıp 1) =
− −
−
e ız(1 e ıp). Hence, eıp = 1 and so p = 2π. The case of sin(z) is similar.
− −
−
(b) We see that sin(z) = 0 if and only if eız = e ız i.e. e2ız = 1. Hence, z = nπ for
−
some n Z.
∈
(c) cos(z) = 0 if and only if e2ız = 1 i.e. 2z = π +2nπ.
−
8
13. (a)
eız +e ız
−
cos(z) =
2
1
= ((ey +e y)cos(x)+ı(e y ey)sin(x))
− −
2 −
1
sin(z) = ((e y +ey)sin(x)+ı(ey e y)cos(x))
− −
2 −
(b)
1
sin(z) 2 = (e2y +e 2y +2sin2(x) 2cos2(x))
−
| | 4 −
1
= (e2y +e 2y +2sin2(x) 2+2sin2(x))
−
4 −
= sin2(x)+sinh2(y)
Similarly for cos(z) 2.
| |
(c) Note that sinh(z)+cosh(z) = ez and cosh(z) sinh(z) = e z. Writing cosh(a+b) =
−
−
1(eaeb + e ae b) (similarly for sinh(a + b)) and using these identities, we arrive at
2 − −
cosh(a + b) = cosh(a)cosh(b) + sinh(a)sinh(b) and sinh(a + b) = sinh(a)cosh(b) +
cosh(a)sinh(b).
(d) The derivative of cosh(z) = ez e−z = sinh(z) and that of sinh(z) = ez+e−z =
−
2 2
cosh(z). Notice that cosh(ız) = cos(z) and sinh(ız) = ısin(z). Hence, cosh2(ız)
−
sinh2(ız) = cos2(z)+sin2(z) = 1.
14. Bydefinition, eız+e−ız = 2andhencee2ız 4eız+1 = 0. Solvingthis,wegeteız = 2 √3.
2 − ±
If z = x+ıy, e yeıx = 4(1 √3). Hence, y = ln(4) and x = π +2kπ where k is an
− 2 ± 2 − ±3
integer.
15. (a) ıı = eıln(ı) = eı(ı(2kπ+π)) = e (2kπ+π) k Z.
2 − 2
∀ ∈
(b) Let b = x+ıy.
ab = ebln(a)
= ebln(a)+ıb(Arg(a)+2πk)
| |
= exln(a) y(Arg(a)+2πk)+ı(yln(a)+x(Arg(a)+2πk))
| | − | |
Hence, infinitely many distinct values appear if and only if either y = 0 (so that the
6
magnitudes are different), or xk is not an integer for infinitely many k, i.e. when x is
irrational (so that the arguments differ by values other than 2πn where n Z).
∈
(c) By (b), firstly y = 0. Secondly, xn = p where p and n are coprime.
16. (a) This means the positive and negative parts of an i.e. pn = an+2|an| and qn = −an+2|an|
don’tconverge(since (p +q ) = a converges, ifoneofthemdidconverge, thenso
n n n
would the other have; But (p q ) = a diverges). Since p and q are positive
n n n n n
P −P | |
P P
9
