Document Text Contents
Page 1
A COLLECTION
OF PROBLEMS ON
COMPLEX ANALYSIS

L. I. Volkovyskii, G. L. Lunts,
. and I.G. Aramanovich
Page 2
A Collection of Problems
on
Complex Analysis
Page 218
SINGULARITIES OF MANYVALUED CHABAOTEB 207
1309. Let
Construct the Riemann surfaces of the functions:
(1) w = y(f(z));
(2) w = Logf(z);
(3) w =Log 1(;)+Logf{:z)·
HINT. First prove that the function /(z) is sohlioht in the disk lzl < I
and has the oirole fzf = I as its natural boundary.
§ 3. Some classes of analytic functions with nonisolated singu
laritiest
Let E be a closed set of points with connected complement !J in the extended
zplane. Let AB, AD and AO be classes of singlevalued analytic functions
in !J, respectively: bounded (the class AB), having a bounded Dirichlet integral
with respect to !J (the class AD), continuous in the extended zplane (the
class AO). If the set Eis in some respect small it can be expected that from
the analyticity of the function in !J and the additional limitations connected
with the class of function, it would follow that the function in E is analytic.
Then the function is a constant in virtue of Liouville's theorem. In this case
the set E is said to be a nul aet of the class of functions considered.
The classes of nul sets corresponding to the given classes of functions are
denoted by NAB• N AD• N AC• and the class of their complements in the
domain !J by 0 AB• 0 AD• 0 AC•
A set Eis said to be .ABremovable in the domain G, containing E, if from
the analyticity and boundedness in GE there follows analyticity everywhere
in G (if G is identical with the extended zplane, then ABremovableness in
Gisequivalent to the fact that EeNAB)· ADremovableness and .AOremova·
bleness of a set E in the domain G are similarly defined (in the case of AO·
removableness, functions are considered of the class AO in G, that is, analytic
in GE and continuous everywhere in G). For brevity F denotes any of the
classes AB, AD, AO, if it is a question of some property common to them.
1310. Prove that the set E with connected complement !J, which
is the set of singularities of a singlevalued function analytic in Q,
is closed.
1311. Prove that the isolated singular points of functions of the
classes .AO, .AB, .AD are removable and the sets of singularities
of the functions of these classes are perfect.
t For this section see L. Alu.FOBS, A. BEtJBLING, (1950), Conformal invariants
and function theoretic nullsets, Acta Math. vol. 83, 100129.
Page 219
208 PROBLEMS ON COMPLEX .ANALYSIS
HINT. For /e.AD consider a.Bf?+0 the Dirichlet integral f J l/'l8 rdrd•
C1<fzi<C11
(a singularity at zero), starting from the Laurent expansion
""
.f(z) = .J: Cnzn •
00
1312. Let G be an arbitrary domain in the extended zplane
and Ek(k = I, 2, ... , q) bounded closed sets with connected com
plements Dk, located in G, no two of which have points in common:
(i#=j).
Prove that the function /(z), singlevalued and analytic in GE
q
E = LJ E", can be represented uniquely in the form
k=l
q
f(z) = ,P(z)+ 21J11i(z),
k=l
where ,P(z) is analytic in G, 1Jlk(z) is analytic in D" and 1J11i(oo) = 0.
In particular, if G is the extended zplane, then
q
J(z) =f(oo)+ L1Pk(z)
k=l
HINT. Separate the sets Et in (} by contours "" (la = I, 2, ..• , q) and in
the domain cut off from the variable domain Gn, converging to G, by the con
tours Yk• represent /(z) by Cauchy's formula..
REMARK, 'Pt(z) has on Ek the same singularities a.s /(z) and is the analogue
of the principal pa.rt of the Laurent expansion in the neighbourhood of an
isolated singularity.
1313. Prove that if Ee NF, then E is .Fremovable in any domain
G => E (removableness of nul sets).
HINT, Make use of the result of the preceding problem.
1314. Prove that if Eke NF (k = I, 2, ... , q) and E1 n EJ = 0
q
(i =F j), then E = LJ Ek c: NF (the property of analyticity of nul
k=l
sets).
1315. Prove that if E c: NF• then E does not contain interior
points.
Page 436
ANSWERS AND SOLUTIONS 425
n
1497. t1(z) =Im f(z) J; oc1cw1c(z), where f(z) conformally maps D onto
k1
the plane with horizontal outs, and
I pi if f (z) = ZiJ + ... • a #: oo,
piz + ... , if a = oo.
n
~ 1 J og(C, z)
1498. t1(z) = 2gg(z, a)+ L.J IXk°'lc(z), w1c(z) = 2k a,;da·
k=l r1c
1499. If /(a) = oo, then the field is formed by the dipole (a; p) where
I' is determined from the expansion off (z) close to the point a;
'( ) 11!!.__ + ... , if a #: oo, z = za
piz+ ••. , if a = oo.
1500. (1) If /(a) == 0, f(b) = oo, then the field is formed by the point
charges (a; 2q), (b; 2q) the flux of the vector intensity through each boundary
contour being equal to irero;
(2) If /(a) = 0, then the field is formed by the point charge (a, 2q),
the flux of the vector intensity through a boundary contour, corresponding to
a circle, in the direction of the outward normal to the domain D being equal
to 4:.irq, end through every other contour being equal to zero;
(3) The field is regular everywhere. The flux of the vector intensity
through boundary contours, transformed into circles, in the direction of the
outward normal to Dis equal to ±4:.irq (+for the contour, transformed into
the outer circle), end through every other contour it is equal to zero.
1501. See problems 1487 and 1489.
n1
1502. (1) t1(Z) = J; or:lc°'1c(z)+c, where the or:1c a.re uniquely determined from
k=l
n1
the system J; Plkott = 2qi (i = 1, 2, ... , n1) (see problem 1104) e.nd c is
k=l
an arbitrary real number. The problem is equivalent to the construction of
the :Bow in D, streamlining the boundary contours I'1c with circulations 4:.irqrc
(k = 1, 2, ... , n), if oo e D, and by the circulations 4:.irq1c (k = 1, 2, •.. , n  1),
4nqn, if oo e D (I'n is the external contour).
(2) t1(z) = t10 (z)2qg(z, a), where t10 (z) is determined as in part (1),
form the charges of the layer 2q1c+2qt, where
!lt = _ _!_ f og(C,a) dB.
2:.ir on
r1c
uoa. "(z) = 2q1 log ~ +c, if !l = 0, and
r1
t1(z) = 2(q1.Aq)log"jZj'"2qg(z,a)+c,
Page 437
426 PROBLEMS ON COMPLEX ANALYSIS
8 (log z+log a)
l 1 2:.iri
where A  og a g(z, a) = lzlA log 1
 loge' 8 (logzloga)
l 2:.iri
for.,,.= (loge)/:.iri, where e < a < 1, if q:;tO (in the notation of problem 1439
the Green's function g(z, a) = Im 4i ( : log 111) for I'= 2:.ir and I'1 = 2:.ir;
the latter from the condition 'P = 0 on the boundary of the ring).
15H. The source (a; q) is transformed into the source (a*; q), where
a• is the point symmetrfoal to a. The function u = 2q log 11
1 +c, where
:ii (111, all
f (111, a) conformally maps the domain D onto the unit disk (here and in what
follows the coefticient of heat conductivity Tc is assumed to be equal to 1).
1505. u  2q log 1 zii I +c. :ii 111a
1506. u = 2~ log1;;~:)1 +c.
sin~+isinh :.irk
q 2a 2a
1507. u = 2_ log +c. •• . m 'sinh nh
s1nra• 2a
1508. u = :Tc log I:~!~~: j +c, e = sn [ ! (111 + ib), u], where Tc is de
. d ..... h l. K 2b termme ..om t e re at1on K = a.
1509. (1) The Green's function g(z, a) of the domain D can be considered
as the temperature created in D by the heat source (a; 2:.ir) when the tempera
ture on the boundary is equal to zero;
n
(2) u(z) = 2q g(z, a)+ 2 u1;w1:(111), where w1(111) is the harmonic
:ii k1
measure of I'1;.
q U1U1 lzl
1510, u = 2 g(z, a)+ 1 ( I ) log +uv where g(z, a) is the Green :ii og r1 ,., f't
function (see the answer to problem 1508).
A COLLECTION
OF PROBLEMS ON
COMPLEX ANALYSIS

L. I. Volkovyskii, G. L. Lunts,
. and I.G. Aramanovich
Page 2
A Collection of Problems
on
Complex Analysis
Page 218
SINGULARITIES OF MANYVALUED CHABAOTEB 207
1309. Let
Construct the Riemann surfaces of the functions:
(1) w = y(f(z));
(2) w = Logf(z);
(3) w =Log 1(;)+Logf{:z)·
HINT. First prove that the function /(z) is sohlioht in the disk lzl < I
and has the oirole fzf = I as its natural boundary.
§ 3. Some classes of analytic functions with nonisolated singu
laritiest
Let E be a closed set of points with connected complement !J in the extended
zplane. Let AB, AD and AO be classes of singlevalued analytic functions
in !J, respectively: bounded (the class AB), having a bounded Dirichlet integral
with respect to !J (the class AD), continuous in the extended zplane (the
class AO). If the set Eis in some respect small it can be expected that from
the analyticity of the function in !J and the additional limitations connected
with the class of function, it would follow that the function in E is analytic.
Then the function is a constant in virtue of Liouville's theorem. In this case
the set E is said to be a nul aet of the class of functions considered.
The classes of nul sets corresponding to the given classes of functions are
denoted by NAB• N AD• N AC• and the class of their complements in the
domain !J by 0 AB• 0 AD• 0 AC•
A set Eis said to be .ABremovable in the domain G, containing E, if from
the analyticity and boundedness in GE there follows analyticity everywhere
in G (if G is identical with the extended zplane, then ABremovableness in
Gisequivalent to the fact that EeNAB)· ADremovableness and .AOremova·
bleness of a set E in the domain G are similarly defined (in the case of AO·
removableness, functions are considered of the class AO in G, that is, analytic
in GE and continuous everywhere in G). For brevity F denotes any of the
classes AB, AD, AO, if it is a question of some property common to them.
1310. Prove that the set E with connected complement !J, which
is the set of singularities of a singlevalued function analytic in Q,
is closed.
1311. Prove that the isolated singular points of functions of the
classes .AO, .AB, .AD are removable and the sets of singularities
of the functions of these classes are perfect.
t For this section see L. Alu.FOBS, A. BEtJBLING, (1950), Conformal invariants
and function theoretic nullsets, Acta Math. vol. 83, 100129.
Page 219
208 PROBLEMS ON COMPLEX .ANALYSIS
HINT. For /e.AD consider a.Bf?+0 the Dirichlet integral f J l/'l8 rdrd•
C1<fzi<C11
(a singularity at zero), starting from the Laurent expansion
""
.f(z) = .J: Cnzn •
00
1312. Let G be an arbitrary domain in the extended zplane
and Ek(k = I, 2, ... , q) bounded closed sets with connected com
plements Dk, located in G, no two of which have points in common:
(i#=j).
Prove that the function /(z), singlevalued and analytic in GE
q
E = LJ E", can be represented uniquely in the form
k=l
q
f(z) = ,P(z)+ 21J11i(z),
k=l
where ,P(z) is analytic in G, 1Jlk(z) is analytic in D" and 1J11i(oo) = 0.
In particular, if G is the extended zplane, then
q
J(z) =f(oo)+ L1Pk(z)
k=l
HINT. Separate the sets Et in (} by contours "" (la = I, 2, ..• , q) and in
the domain cut off from the variable domain Gn, converging to G, by the con
tours Yk• represent /(z) by Cauchy's formula..
REMARK, 'Pt(z) has on Ek the same singularities a.s /(z) and is the analogue
of the principal pa.rt of the Laurent expansion in the neighbourhood of an
isolated singularity.
1313. Prove that if Ee NF, then E is .Fremovable in any domain
G => E (removableness of nul sets).
HINT, Make use of the result of the preceding problem.
1314. Prove that if Eke NF (k = I, 2, ... , q) and E1 n EJ = 0
q
(i =F j), then E = LJ Ek c: NF (the property of analyticity of nul
k=l
sets).
1315. Prove that if E c: NF• then E does not contain interior
points.
Page 436
ANSWERS AND SOLUTIONS 425
n
1497. t1(z) =Im f(z) J; oc1cw1c(z), where f(z) conformally maps D onto
k1
the plane with horizontal outs, and
I pi if f (z) = ZiJ + ... • a #: oo,
piz + ... , if a = oo.
n
~ 1 J og(C, z)
1498. t1(z) = 2gg(z, a)+ L.J IXk°'lc(z), w1c(z) = 2k a,;da·
k=l r1c
1499. If /(a) = oo, then the field is formed by the dipole (a; p) where
I' is determined from the expansion off (z) close to the point a;
'( ) 11!!.__ + ... , if a #: oo, z = za
piz+ ••. , if a = oo.
1500. (1) If /(a) == 0, f(b) = oo, then the field is formed by the point
charges (a; 2q), (b; 2q) the flux of the vector intensity through each boundary
contour being equal to irero;
(2) If /(a) = 0, then the field is formed by the point charge (a, 2q),
the flux of the vector intensity through a boundary contour, corresponding to
a circle, in the direction of the outward normal to the domain D being equal
to 4:.irq, end through every other contour being equal to zero;
(3) The field is regular everywhere. The flux of the vector intensity
through boundary contours, transformed into circles, in the direction of the
outward normal to Dis equal to ±4:.irq (+for the contour, transformed into
the outer circle), end through every other contour it is equal to zero.
1501. See problems 1487 and 1489.
n1
1502. (1) t1(Z) = J; or:lc°'1c(z)+c, where the or:1c a.re uniquely determined from
k=l
n1
the system J; Plkott = 2qi (i = 1, 2, ... , n1) (see problem 1104) e.nd c is
k=l
an arbitrary real number. The problem is equivalent to the construction of
the :Bow in D, streamlining the boundary contours I'1c with circulations 4:.irqrc
(k = 1, 2, ... , n), if oo e D, and by the circulations 4:.irq1c (k = 1, 2, •.. , n  1),
4nqn, if oo e D (I'n is the external contour).
(2) t1(z) = t10 (z)2qg(z, a), where t10 (z) is determined as in part (1),
form the charges of the layer 2q1c+2qt, where
!lt = _ _!_ f og(C,a) dB.
2:.ir on
r1c
uoa. "(z) = 2q1 log ~ +c, if !l = 0, and
r1
t1(z) = 2(q1.Aq)log"jZj'"2qg(z,a)+c,
Page 437
426 PROBLEMS ON COMPLEX ANALYSIS
8 (log z+log a)
l 1 2:.iri
where A  og a g(z, a) = lzlA log 1
 loge' 8 (logzloga)
l 2:.iri
for.,,.= (loge)/:.iri, where e < a < 1, if q:;tO (in the notation of problem 1439
the Green's function g(z, a) = Im 4i ( : log 111) for I'= 2:.ir and I'1 = 2:.ir;
the latter from the condition 'P = 0 on the boundary of the ring).
15H. The source (a; q) is transformed into the source (a*; q), where
a• is the point symmetrfoal to a. The function u = 2q log 11
1 +c, where
:ii (111, all
f (111, a) conformally maps the domain D onto the unit disk (here and in what
follows the coefticient of heat conductivity Tc is assumed to be equal to 1).
1505. u  2q log 1 zii I +c. :ii 111a
1506. u = 2~ log1;;~:)1 +c.
sin~+isinh :.irk
q 2a 2a
1507. u = 2_ log +c. •• . m 'sinh nh
s1nra• 2a
1508. u = :Tc log I:~!~~: j +c, e = sn [ ! (111 + ib), u], where Tc is de
. d ..... h l. K 2b termme ..om t e re at1on K = a.
1509. (1) The Green's function g(z, a) of the domain D can be considered
as the temperature created in D by the heat source (a; 2:.ir) when the tempera
ture on the boundary is equal to zero;
n
(2) u(z) = 2q g(z, a)+ 2 u1;w1:(111), where w1(111) is the harmonic
:ii k1
measure of I'1;.
q U1U1 lzl
1510, u = 2 g(z, a)+ 1 ( I ) log +uv where g(z, a) is the Green :ii og r1 ,., f't
function (see the answer to problem 1508).