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Section 10.2:
4. Follows from the de¬nition and (5.18) (p. 90).
6. Consider the operator T = T1 — . . . — Tq , and use Exercise 2 on p. 161.


Section 10.3:
˜
2. Let Tj = T1 — . . . — Tj and use the previous exercise, with a suitable
choice of projections.


Section 10.4:
ˆ
1. Apply Theorem 50 to f (z p ).
ˆˆ
2. Note that h2 = S (Bk2 f2 ) is holomorphic and of exponential growth at
most k2 in some S(d2 , µ). Therefore, g = g1 + g2 , with g1 holomorphic
at the origin, and g2 = Lκ holomorphic in S(d2 , µ+π/κ). This implies
ψ(z) = g2 (z) ’ g2 (ze2πi ), hence ψ is holomorphic in S(d2 ’ π, µ +
π(1/κ ’ 2)).
3. Using Exercise 1 on p. 99, show the existence of ψ(z), holomorphic in
S = S(d2 ’ π, µ + π(1/κ ’ 2), r), having no holomorphic continuation
beyond |z| = r, and so that ψ(z) ∼1/κ ˆ in S. De¬ne f (z) = fn z n ,
ˆ
0
=
a
fn = “(1 + n/k1 ) 0 ψ(w)w’n’1 dw, n ≥ 1, for some a ∈ S, and use
the previous exercise.
q

4. Consider convergent series gj with
ˆ j=1 gj
ˆ 0.


Section 10.5:
2. Use Theorem 50 and Exercise 5 on p. 73.


Section 10.6:
1. Argue as in the proof of Theorem 37 (p. 106).
2. Use Proposition 22 (p. 168) and Proposition 13 (p. 105).
262 Solutions

Section 10.7:

gn z n ∈ C {z}1 . The singular
1. Note fn = “(1 + n/2) gn , and show
multidirections d = (d1 , d2 ) are those with d2 = 0 modulo 2π. Observe
that, owing to our identifying certain singular directions, these are
indeed ¬nitely many.
1/2
3. For the series in Exercise 1, show h± (z) = 0 ψ(w) (w ’ z)’1 dw,
for 0 < ± arg z < µ, so their di¬erence equals ψ(z).


Section 10.8:
ˆ ˜
1. Show that d = (d1 , . . . , dq ) is nonsingular for f if and only if d =
(d1 ’ 2π/p, . . . , dq ’ 2π/p) is so for g .
ˆ

2. Show p fj (z p ) = z ’j
p’1 ’j ˆ
ˆ
=0 µ f (µ z), and use the previous exercise,
together with Theorem 51 (p. 166).

ˆ
3. Let k be the optimal type of summability for f . If the correspond-
ing values κj are all larger than 1/2, apply Theorem 50 (p. 164).
Otherwise, introduce additional kj ™s so that then the theorem can be
applied.


Section 11.1:

2. Substituting t = x1/± in the previous exercise, show the representa-
tion
∞(d)
±
t±’1 e’t ezt dt, ’π/(2±) < d < π/(2±).
D± (z) = ±
0




Section 11.3:
1 1
1. Show 0 h(z xs ) dx = 0 f (z (1 ’ x)s ) g(z xs ) dx for h = f —k g and
s = 1/k, and let s ’ 0.


Section 12.1:

3. Show km (z) = (1 ’ z eam (z’1) )/(1 ’ z).
Solutions 263

Section 12.4:
˜
1. Determine P (z) and E(z) as follows: Set e1j (z) = e1j (z), 1 ¤ j ¤ ν.
˜
Then not all e1j (z) can vanish at the origin, so assume e11 (0) = 0;
˜ ˜
˜
otherwise, permute the columns of E(z) accordingly. Now, assume
˜
that the entries of P (z) and E(z) in rows with numbers ¤ j ’ 1 are
known, and that eµµ (0) = 0, 1 ¤ µ ¤ j ’ 1 (which is correct for
˜
j = 2). Then for 1 ¤ < j, we can recursively determine pj (z) by
requiring that ej (z) ’ µ=1 pjµ (z) eµ (z) = O(z k ’kj +1 ); because of
˜
e (0) = 0, this determines pj (z) uniquely. Setting
˜
j’1
ej (z) = ej (z) ’
˜ pjµ (z) eµ (z),
˜
µ=1

1 ¤ ¤ ν, the ¬rst j ’ 1 entries vanish as required, while of the
remaining ones at least one cannot vanish at the origin. By a permu-
tation of the columns with numbers ≥ j we then can arrange that
ejj (0) = 0.
˜
2. Factor E(z) as in the previous exercise, and observe that z K P (z) z ’K
is an analytic transformation, while z K E(z) z ’K is entire.
˜

Section A.1:
1. x11 = x22 = 0, x12 = x21 = 1/2.
2. Observe J2 = 0, resp. J1 = 0, in case j = 1, resp. k = 1.
˜
3. Show ¬rst that » is an eigenvalue of A if and only if a nonzero X ∈
C k—j exists satisfying A X ’ X B = »X. Then, use Lemma 25 to
˜
show that » being an eigenvalue of A implies » = »1 ’ »2 , with »1
being an eigenvalue of A and »2 one of B. Finally, consider matrices
X of the form X = x · xT to conclude that every such » is indeed an
˜
˜
eigenvalue of A.

Section A.2:
1. If A11 or A22 are not invertible, ¬nd a suitable vector c = 0 with
A c = 0, if they are, verify the given formula for A’1 .

Section A.3:
1. Consider the power series expansion of φ about z0 to show |φ(z)| ¤
c |z ’ z0 |2 for |z ’ z0 | ¤ µ, with su¬ciently small µ > 0 and su¬ciently
large c ≥ 0.
2. Show |φ(z)| ¤ (1 ’ 1/n) |z ’ z0 | + c |z ’ z0 |2 , with µ and c as above.
264 Solutions

Section B.1:
1. Use linearity of φ to show

φ[(z ’ z0 )’1 (f (z) ’ f (z0 ))] = (z ’ z0 )’1 [φ(f (z)) ’ φ(f (z0 ))],

and then let z ’ z0 , using continuity of φ.
2. Observe
T (z) f (z) ’ T (z0 ) f (z0 ) T (z)[f (z) ’ f (z0 )] [T (z) ’ T (z0 )]f (z0 )

= ,
z ’ z0 z ’ z0 z ’ z0
and analogously for ± in place of T .

Section B.2:

1. Assume f (z) = m fn (z ’ z0 )n , for some m ≥ 0, and set g(z) =
(z ’ z0 )’m f (z). Then g is holomorphic in D(z0 , ρ), and g(zk ) = 0.
Letting k ’ ∞ implies g(0) = fm = 0.
2. Use Cauchy™s integral theorem and proceed as in case of E = C .
5. Use Theorem 27 to show that all Taylor coe¬cients fn , for n ≥ 1, are
equal to zero.

Section B.3:
1. Use holomorphic continuation by successive re-expansion.
2. Use the previous exercise.
3. Use the previous exercises together with Exercise 1 on p. 223.
4. For (a), use (B.8) and a change of variable t = x u. For (d), observe
that we have shown the left-hand side of (B.12) to have a limit; to
evaluate it, it su¬ces to consider positive real values of z.
5. Show (1 + a/z)a+z = exp[(a + z) log(1 + a/z)] ’ ea for |z| ’ ∞ in
sectors | arg z| ¤ π ’ µ.
6. Observe that the left-hand side is “(1 + z) “(1 + n) nz /“(1 + n + z)
and use the previous exercise.
10. Justify interchanging the order of integration and use Cauchy™s for-
mula to evaluate the integral. Then, prove (B.14) for Re z > 0, and
use Exercises 3 and 7 to show the same for general z ∈ C .
11. Use the identity theorem of Exercise 3 to see that the formula (B.14),
proven for Re z < ’1, extends automatically.
Solutions 265

Section B.4:
1. Under the assumptions made, there is exactly one w on the path γ
for which w± = z.
2. The path γ may be replaced by one, for which the two radial parts,
instead of the negative real axis, are along the rays arg w = ±(µ +
π/2), for arbitrarily small µ > 0. Doing so, the denominator in the
integral never vanishes.
3. Use Theorem 69 (p. 233), together with Stirling™s formula in Theo-
rem 68 (p. 229).

Section B.5:
1. Study the function f (z) = exp[a z k ], a ∈ C .
2. Either f is constant or of order larger than π/(β ’ ±).

Section C.1:
1. Use Lemma 24 (p. 212) to show the existence of T so that
I 0 A11 0 I 0
A= ,
’T
T I 0 A22 I
and then use the rules for the exponential of a matrix.
2. If N is a ν — ν Jordan block, then
® 
1 0 ... 0 0
 log z 
1 ... 0 0
 
N
z = .
. . . .
..
° »
. . . .
.
. . . .
ν’1 ν’2
log z log z . . . log z 1

3. Verify that it su¬ces to consider A in Jordan canonical form, and then
use the rules for computing exponentials of diagonal, resp. nilpotent,
matrices.
4. Comparing the quadratic terms in the power series expansions of both
sides, one ¬nds as a necessary condition that (A+B)2 = A2 +2 A B +
B 2 , which holds if and only if A and B commute.
5. By induction, show
d
n ≥ 1,
[A(z)]n = n [A(z)]n’1 A (z) = n A (z) [A(z)]n’1 ,
dz
[A(z)]n /n!
and then justify termwise di¬erentiation of
266 Solutions

Section C.2:

1. A = T ’1 diag [0, 2πi] T , hence eA = T ’1 diag [e0 , e2πi ] T = I.

2. By de¬nition, branches of the matrix logarithm cannot have eigen-
values that di¬er by integer multiples of 2πi.

3. Insert into (C.1) and justify termwise integration.
References




[1] L. V. Ahlfors, Complex Analysis, McGraw“Hill, New York, 1966.
[2] D. V. Anosov and A. A. Bolibruch, The Riemann“Hilbert Prob-
lem, Vieweg, 1994.
[3] D. G. Babbitt and V. S. Varadarajan, Formal reduction of

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