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239 views18 pagesChapter1 Solution
Complex Analysis Chapter 1 solution
Uploaded by
Chengyuan XiangCopyright
© © All Rights Reserved
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/ 18
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Chapter1
Preliminaries to Complex Analysis
Berret
July 10,2020
Abstract
‘This chapter is devoted to the exposition of basic preliminary material
‘we uso extonsively throughout of this book. I will help you salve serval
‘exercises in this chapter, thore may be some spelling mistakes or even wrong,
methods,
part to the inroducton of complex mers; paredoreally, thi eased on the soamingly
Timi anton Unt Uhre are ashe ems ean ore ptr
B, Borel, 1052
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Exercise 1 is very easy, so we a
it and begin with Exercise 2
Exercise 2. Let <,> denote the usual inner product in R®. In other word
8, if 2 = (21,44) and W = (22,49), thon
=m + yy
Similarly, we may define a Hermitian inner produet (+) in € by
(=,w) = 20
‘The term Hermitian is used to deseribe the fact that (.,-) is not symmetric, but,
rather satisfies the relation
(w)=20 for all zwet.
Show that %
<2,W> = 5l(2 1) + (12)] = Pele),
Where we use the usual identification 2 = -+ iy € C with(x, ) €R?,
Solution 2. Though this is a straightforward calculation, ut we have two ways
to solve it
We suppose that z = 21 +iz2,w = wy + iu, 80
= Re ((z21 +22) (w1 — iw) = 2yw1+z909
Wayl
[(cr-bize) wy —iwy)-+(21—i22)(w1-bins)] = zywy-b22t0e
Way2
1
gllem)+e,2)
1
ple) +w.2)
Exercise 3. With w = se” , where s > 0 and y € R, solve the equation
2" =w in C where n is a natural number. How many solutions are there?
Solution 3. We suppose that z= te’, whore t > 0 as s > 0 and n is a natural
umber. Then we have
Pe? = geil
Wo easily find that 2 = ste8+28", whore k= 0,1,+
nith root of s.
In summary, there are n solutions as has n values,
n—1 and sm is the real
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:
(0) For any two complex numbers 2,1, one and only one of the following is true:
zh ww > 2orZ =u.
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For all 21,22, 23 € € the relation 2) > 22 implies 21 425 > 22425,
)) Moreover, for all 21,225 23 € € with =3 > 0, then 21 > 22 implios 2123 >
28s,
Solution 4. We can conclude that ¢=0. If not, just for a contradiction, that
i» 0, then —1 =i-i » 0-7 =0. Now we may suppose that 0 > i, but also.
similarly, 0= 0-1 + 4-7 =—1 . So we must have i =0. But then for all z €C we
have 2-i= 2-0 =O. So this relation would not give a trivial total ordering.
Exercise 5 and Exerc
tum to Exercise 7.
is not so easy to write down, so I skip them and we
Exercise 7. The family of mappings introduced here plays an important role
in complex analysis. ‘These mappings, sometimes ealled Blaschke factors, will
reappear in varions applications in later chapters.
(a) Let 2,w be two complex numbers such that Zw # 1.Prove that
if [2] < lando| <1,
‘and also that
Poel lel = tor fu] 1
(b) Prove that for a fixed w in the unit dise D, the mapping
satisfies the following conditi
(i) F maps the unit dise to itsolf(that is, F : D> D),and is holomorphic.
(Gi) F interchanges 0 and w, namely (0) = w and F(w) =0.
(iii) [F(2)] = 1 of [2]
(iv) FD > D is bij
Solution 7.
(a) We have two ways to solve it.
[Wayl[lt is a straightforward ealculation.
We suppose that z = = 4 iz2,w — w1 + wp, thon to calculate
to caleulate eZ
wie
‘We noed to compare it with 1,
that is, to compare wy — 21 + (wa — 29) with 1 — wrt — wo22 + i(wizg —wrzt),
that is, to compare (ws — 21)? with (1 —wy21 — wy22)? + (wi22 — wr2)?,
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that is, to compare jto|? + |2|? —2(w121 + wa2a) with ||? |2[ 41 —2(w.21 + w222),
that is, to compare |t[” + |2[? with Jw]? J2|? +1.
When Jw] < 1,2] <1, (1 —|wl?)(1— 2/7) > 0.
So we know that
O and —m <0< x.
Solution 9. A straight caleulation can solve it.
We know that z= 2-4 iy,r = peosip,y = psing
then
vate. yp, 2!
$2) = lay) +a). T= Fe By
du_ du Br Ou By _ ou
Dp de Dp * By Op Oe
Similarly, we also have
0 we have
oe
a
dv _ bv Be, Oy dy OW ss
Op dx dp” dy Ip dz “P+ |, HP
du du Or Gu By du ou
Ben Be Bet Oy Be 7 ae CPsne) +5 ping
serene
om
a
sf)
Jogz is holomorphic in the region r > 0 and —« <0.<4
Exercise 10. Show that
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where A is the Laplacian
Solution 10. Tb prov we shall begin with
ao om ve
Dr” Oz Dr” BF De Oz” OF
aa a
Oy” a OF
eo, FF oe
02” d202" a8” Fz
PP Po
o_o
ae aed:
te
‘Bete * obs
so lastly we get bo 5
Oabs ~ “oes
Exercise 11. Use Exercise 10 to prove that if f is holomorphic in the open set ,
then the real and imaginary parts of J are harmonic; that is, their Laplacian is
Solution 11.
0- (a*e) ae (55) ay (oe) ~ aay apts
em (Bisa) °~ a: ay) ay (ae) ~ aa *
14 iv is a holomorphic fimetion and 1, is continuous in the defined area.
‘Thos,
eo tu Pu Oe
Ody” OyOr * Drdy~ Bye
naginary parts of fis harmonic; that is, their Laplacian: Au, Aw
So the real and
Exercise 12. Consider the function defined by
f(x +iy) = Vicilyl,where 2,yeR.
Show that f satisfies the Cauchy-Riemann equations at the origin, yet f is not
holomorphic at 0.
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Solution 12. We ean eas
vis continnousty
ferentiablo,
At the origin, $f = $ = 0, and as v = 0, #2 = 3 = 0, which means that f
satisfies the Caichy-Riemann equations at the origin, Vet. fis not holomorphic at 0
got that w— //fa][yl and » 0
rentiable while u is differentiable but not conti
ously
Exercise 13. Suppose that f is holomorphic in an open set 2. Prove that in
any one of the following eases:
(i) Re(f) is constant
Gi) Im (J) is constant
(Gi) [7] is constant
‘one ean conelude that f is constant.
Solution 13. Let f(z) =f (x,¥)
‘u (x,y) + iv (x,y), where 2 = 2+ iy.
() Since Re (f) is constant,
as we know that f is holomorphic in an open set (2, by the Cauchy-Riomann
equations,
av
a
then in 2, 4
reo- 2-3
thus /(2) is constant.
(i) Since Im(f) is constant,
by the Cauchy
thus f(2) is constant,
(ili) We first give a mostly correct. argument; the reader should pay attention
to find the difficulty. Since |f| = /2?-+ 4? is constant, we have
awe +02)
o= ee)
ati? +02)
0
1
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Plug in the Cauchy-Riemann equations and we get
thus we have
BeBe
dr udy
w+ av
uo Oy
which means u? +0? = 0 or 8 = 0. If u? +0? =0, then, since 1,0 are real,
6, and thus f = 0 which is constant. Thus we may assume w? + v? equal
‘a non-zor0 constant, and we may divide by it. We multiply both sides by w and
find $2 =, plug back in the early equation, we get $2 = 0, by Canchy-Riemann
equations, 8 — 0
£@
Oe "De
af du | ov
Oe
Thus f is constant.
Why ix the above only mostly a proof? ‘The problem is we have a division by
4, and need to make sure everything is well-defined. Specifically, we nced to know
that n is never zero. Wo 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
au? se) uy ov
9 — SY yy BH 5 yt
0 = A A au 4
hug in the Cauchy-Riomann equations and we got
bv
oe)
402
on Oy
‘Wo amtiply the first equation 1 and the second by 1, and obtain
Ou ae
ing the two yields
22h 4 2e
oy OF
8
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‘manner as before, except now we don’t have the ane
noving 1 in the denominator. If12-+1 = 0 then 0, else we enn divide by
u? 4-0? =O and find 2¢ = 0, Arguing along these lines finishes the proof.
One additional remark by Jeff Meng: we ean trivially pass from results
‘on patrials with respect to v to those with respect to u by noting that if f = u+iv
thas constant magnitude, so too docs g — if — —v-+ iu, which essentially switches
the roles of u and u, Though this isn’t noeded for this problem, arguments such
as this ean be very nseful.
‘The following is from Steven Miter. Let’s consider the proot. If|f|=0
the problem is trivial as then f =0, so we assume || equals a non-zero constant.
As |f| is constant, |f|? = ff is constant, By the quotient rule, the ratio of two:
holomorphic fanctions is holomorphic, assuming the denominator is non-zero. We
thus find f = 4 is holomorphic. ‘Thus f and f are holomorphic, and satisfy the
Canchy-Riomaiin equations. Applying these to f = ui + iv yields
ou oe
Oe” dy dy Oe
while applying to J =u + i(—v) gives
au _ a0) Oe
Or ~ Oy * Oy
Adding these equations together yields
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
theorom, the solution is similar to Real Analysis, so we skip and just retell them.
Theorem 9. (summations by parts) Suppose {ay}¥_, and {bn} are two
finite sequences of compler numbers. Let By = *_,b, denote the partial sums
of the series by with the convention By —0, then
n wa
Y tal = andy —anbar = So (@up1 =a) Br
i oH
Theorem 3. (Abel’s theorem) Suppose I~ aq converges, then
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As to Exereise 16,17, we use Hadamard’s formula and directly solve them. Ac-
tually, 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
hhas a power series expansion around any point. in its disc of convergence,
Solution 18. We suppose
B02) = Fags
and or any 2 £9, then write z= 2+ ee 2), a8 we know that f(z ) converges,
Sento +e- ar
“ Soule char a
-Fae-ay"
tere
DC satensi
and b,, actually also converges.
Exercise 19. Prove the following:
(i) The power series J>n2" does not converge on any point of the unit circle.
(Gi) The power series "/n? converges at any point of the unit circle.
(iii) The power series <2" /n converges at every point of the mit circle except
zal.
Solution 19.
(i) On the unit circle, we suppose = = cos +isind, 0 € [0,2n), then ay =
neos8 + insind
We know that lima sc neosné doesn't exist ¥@ € (0,2). But when 0 =
have lity 400 8 6080 = +00, 80 lilly» 420 My JOESHE exist LOO.
Now we can deduce that S755 a, diverges, that is, ‘The power series m2"
does not converge on any point of the unit circle.
(Gi) On the unit circlo, we suppose z = cosd + isin#, 0 € [0,2n), then a4 =
2"/n? = cosnd/n? + isin dn?
As SOE 1 db converges, so EXP ay converges.
10
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Sinnilarly,
by, converges.
‘also converges as to be proved above.
‘we suppose 2 = cos + isin®, @ € [0,2r), then
“+ iby
that is, when 2 = 1, O45; 22 = OX, + obvionsly diverges
When @ € (0,2n), 2%" and = 226" hoth converges by Dirichlet dis-
For the integrity of the proof, I will explain it briefly.
As.can be easily proved, 3 cos n6’s partial sum sequence is Spe 3, which
is bounded when 0 € (0,2n), {2} is monotonically decreasing and tending to zero,
so 552%" converges. Similarly, 8"? also converges.
simmary, the power series $7 2"/n converges at every point of the un
except z= 1.
(iii) On the unit
soul
rele
Exercise 20. Expand (1 — z)-" in powers of z. Here m is a fixed positive
jeger. Also, show that
a2" = Sanz
‘then one obtains the following asymptotic relation for the coefficients:
1 ma
Oa Genin | a nee
Solution 20. Directly use Taylor's formula at zero, we get
where
(m+n—1)!
on = “tm — Din!
‘To prove 1
On Gay a Be
wo just need to prove
(mint do
“fa Dat ~ fay he
that is to prove
(n41)-(m4n—1) wnt
i
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‘The number of items in the left formula is actually m —1, thus we just need to
prove
m=1
~d
a4 da42.a4
We already learn that m is a fixed positive integer and m+ 00, thus the number
of items in the left formula is limited, we can directly prove it.
Exercise 21. Show that for [2] <1, one has
a
and : om
2, de ake
tee tia treat
Tustify any change in the order of summation,
Solution 21. Por the first formula, a calculation can solve it.
ke Dea
12
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for |2| <1, the formula equals 0, which moans
a2 ee
1+z° 142 142
There must be some concise solutions, but the solution here
out.
very easy to come
Exercise 22. Let N= {1,2,3,-+-} denotes the set of positive integers. A subset
SC Nis said to be im arithmetic progression if
$= {aa dat Yat dd}
where a,d €N. Here d is called the step of 8.
Show that N cannot be partitioned into a finite number of subsets that ar
arithmetic progression with distinct steps (except for the trivial ease @ = d= 1).
Solution 22. Suppose S,,a = {4,0 +d,a-+2d,a+3d,---}
Yo eet et et cg atte a tn 2 (siete)
let |z| <1, we have
y
‘at the same time, we have
nen *
‘To find a finite number of subsets with different elements and steps d,we can just
pick by
42,6,10,--- 5
80 we ean got
w- Ot DO tee Ds
net eSagtn
142"
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‘To any subsets with fini
umbers, such 18 Sq, Soa.dgy*"* Sanda
Da < him So —
So to subsets with finite numbers, if a and d cannot be 1 at the same time, it’s
impossible that > Said QED.
Exercise 23. Consider the funetion f defined on R by
oO if r<0
ro-{ eve if e>0
Prove that f is indefinitely differentiable on R, and that f)(0) =0 for all n > 1
Conclucie that f does not havo a converging power series expansion °° y a2"
for a near the origin.
Solution 23. When x € (-00,0], f(x) =0 is obvious. So it’s derivable of any
order and
tin
0 f
When x € (0,400), n = 1, f'(e) = <2e~# Use mathematical methods of
Induction, then suppose when n
sa)-e# PQ), Pe)=-Te
then when n= k-+1, we ean deduce that
seck re 2
ak [Pe - are]
-e)
whore 1
Q@)=Log
and
ee
noo a” =0
Obviously, f(2)'s derivable of any order of kean be expressed as 6 #-P(x), where
P(2) is a polynomial function of . And when x © (0,400), P(z) is derivable
everywhere, so J(2) ean be derived at any order.
“Then in a suuall neighborhood (0,4), f(e) = C29 4
As to
1
=”
1
ont
(ay
4
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Vn, M > 0, we can find 2, when » satisfies
ay|>™
we have
80 the series does not converge near the origin,
Exercise 25. ‘The next three calewlations provide some insight into Canchys
tcorem, which we treat in the next chapter
(i) Braluate the integrals
[ew
for all integers n. Here + is any circle centered at the origin with the positive
(counterclockwise) ori
(i) Same question as before, but with 7 any circle not containing the origin.
(iii) Show that if Jal