No exercises are assigned in this section.
Chapter 5
Singularities
5.1 Classification of singularities
Exercise 1. Each of the following functions f has an isolated singularity at z = 0. Determine its nature; if it is a removable singularity define f (0) so that f is analytic at z = 0; if it is a pole find the singular part; if it is an essential singularity determine f ({z : 0 < |z| < δ}) for arbitrarily small values of δ.
a)
f (z) =sin z z ; b)
f (z) = cos z z ; c)
f (z) =cos z− 1
z ;
d)
f (z) = exp(z−1);
e)
f (z) = log(z + 1) z2 ; f)
f (z) = cos(z−1) z−1 ; g)
f (z) = z2+ 1 z(z− 1); h)
f (z) = (1− ez)−1; i)
f (z) = z sin1 z;
j)
f (z) = znsin1 z.
Solution. a) According to Theorem 1.2 p.103, z = 0 is a removable singularity, since
zlim→∞(z− 0)sin(z)
then f is analytic at z = 0. A simple computation using L’Hospital’s rule yields limz→0sin(z) z = 1.
z + analytic part.
Thuscos zz has a simple pole at z = 0.
c) According to Theorem 1.2 p.103, z = 0 is a removable singularity, since
zlim→∞(z− 0)cos(z)− 1
then f is analytic at z = 0. A simple computation using L’Hospital’s rule yields limz→0cos(z)z−1 = 0.
e) We know So we can write
log(1 + z) Definition 1.8 p. 105).
g) First, we simplifyzz22+1
−z = 1 +z(zz+1
−1). Now partial fraction decomposition of z(zz+1
−1) yields
and hence f (z) =zz22+1
−z has poles of order 1 at z = 0 and z = 1 and the singular part isz2
−1−1z (by Equation 1.7 and Definition 1.8 p. 105).
i) We know (see p. 38)
Therefore f (z) = z sin1z has an essential singularity at z = 0 (it is neither a pole nor a removable singularity by Corollary 1.18c). We have
f ({z : 0 < |z| < δ}) = f [ann(0; 0, δ)] = C
by the Casorati-Weierstrass Theorem p. 109. But f ({z : 0 < |z| < δ}) = f [ann(0; 0, δ)] is either C or C\{a}
for one a∈ C by the Great Picard Theorem (p. 300). In this case, f ({z : 0 < |z| < δ}) = f [ann(0; 0, δ)] = C.
Exercise 2. Give the partial fraction expansion of r(z) = (z2+z+1)(zz2+1−1)2. Solution. Not available.
Exercise 3. Give the details of the derivation of (1.17) from (1.16).
Solution. Not available.
Exercise 4. Let f (z) = z(z 1
−1)(z−2); give the Laurent Expansion of f (z) in each of the following annuli: (a) ann(0; 0, 1); (b) ann(0; 1, 2); (c) ann(0; 2,∞).
Solution. a) We have 0 <|z| < 1 and using partial fraction decomposition yields
f (z) = 1
c) We have 2 <|z| < ∞ and thus n. Determine the singular part of f at each of these poles.
Solution. Define zn= π
2 + nπ. We know tan(z) = cos(z)sin(z) and (cos(z))0=− sin(z). Hence, tan(z) has a simple pole at each znwith residue−1 and with singular part −z−z1n at each zn.
Exercise 6. If f : G→ C is analytic except for poles show that the poles of f cannot have a limit point in G.
Solution. Not available.
Exercise 7. Let f have an isolated singularity at z = a and suppose f (z) . 0. Show that if either (1.19) or (1.20) holds for some s in R then there is an integer m such that (1.19) holds if s > m and (1.20) holds if s < m.
Solution. Let limz→a|z − a|s| f (z)| = 0 hold, that is limz→a(z− a)sf (z) = 0 for some s in R. Then it holds obviously for any greater s’s. Hence, there exist an integer n > 0 such that
limz→a(z− a)nf (z) = 0 (5.1)
and
limz→a(z− a)n+1f (z) = 0 (5.2)
(In fact we are free to choose any positive integer greater than s). Now, (z− a)nf (z) thus has a removable singularity at z = a (by Theorem 1.2 p. 103) applied to (5.2)) and its extended value is also zero by (5.2).
Thus (z− a)nf (z) has a zero of finite order k at z− a. So
where m = n− k.
Now let limz→a|z − a|s| f (z)| = ∞ for some s in R. Then there exists an integer n < 0 so that
limz→a|z − a|n| f (z)| = ∞ (5.3)
(In fact we are free to choose any negative integer less than s). Now (z− a)nf (z) thus has a pole at z = a (by Definition 1.3 p.105 and (5.1)) and (z− a)nf (z) has a pole of finite order l at z = a and so
(z− a)nf (z) = (z− a)−lk(z) where k(z) is analytic at a and k(a) , 0 (Proposition 1.4 p.105). Hence
limz→a(z− a)sf (z) = lim
z→a(z− a)s−n+nf (z) = lim
z→a(z− a)s−n−lk(z) =
0, s > n + l
±∞, s < n + l k(a) , 0, s = n + l.
Thus,
limz→a|z − a|s| f (z)| =
0, s > m
∞, s < m where m = n + l.
Exercise 8. Let f , a, and m be as in Exercise 7. Show: (a) m = 0 iff z = a is a removable singularity and f (a) , 0; (b) m < 0 iff z = a is a removable singularity and f has a zero at z = a of order−m; (c) m > 0 iff r = a is a pole of f of order m.
Solution. a)⇒: Let m = 0. Then by Exercise 7, we have
limz→a(z− a)2f (z) =
0, s > 0 = m
,0, s = 0 = m. (5.4)
Thus using s = 1, f (z) has a removable singularity (by Theorem 1.2 p. 103) with the extended value equal to f (a) = limz→a f (z) , 0.
⇐: Let z = a be a removable singularity and f (a) = limz→a f (z) , 0. Then limz→a(z− a) f (z) = 0 by Theorem 1.2 p.103. By Exercise 7, we obtain that there is an integer m such that
limz→a|z − a|s| f (z)| =
0, s > m
∞, s < m. But m = 0, since
limz→a(z− a)sf (z) =
0, s > 0
∞, s < 0, which follows by (5.4).
b)⇒: Let the algebraic order be defined at z = a and be equal to −m (m > 0). Then by Exercise 7, we have limz→af (z) = 0 and lim
z→a(z− a) f (z) = 0.
Recall limz→a(z− a)sf (z) = 0 if s > −m so it is true of s = 0 and s = 1. Thus, f (z) has a removable singularity at z = a and also the extended function has the value limz→a f (z) = 0 at z = a (p. 103 Theorem
1.2). Now h(z) := (z− a)−mf (z) has a removable singularity at z = a (since limz→a(z− a)−m+1f (z) = 0) but h(z) has extended value
h(a) = lim
z→a(z− a)−mf (z) , 0
(by Exercise 7). Therefore f (z) = (z− a)mh(z) has a zero of order m at z = a.
⇐: Assume f (z) has a removable singularity at z = a and let z = a be a zero of order m > 0. Then f (z) = (z− a)mh(z) where h(z) is analytic in|z − a| < δ, δ > 0 and h(a) , 0. Then
limz→a|z − a|s| f (z)| = limz
→a|z − a|s+m|h(z)| =
0, s >−m
∞, s < −m by Exercise 7. Hence, the algebraic order of f at z = a is equal to−m (m > 0).
c)⇒: Let the algebraic order be m > 0. Then
limz→a(z− a)sf (z) = 0 for s > m by Exercise 7, that is
limz→a(z− a)m+1f (z) = 0 and lim
z→a(z− a)mf (z) , 0
by Exercise 7. Then for z , a, we have h(z) = (z− a)mf (z) has a removable singularity at z = a with h(a) , 0. Now f (z) = (z− a)−mh(z) where h∈ A(B(a; δ)) and h(a) , 0. Thus f (z) has a pole of order m at z = a.
⇐: Let f (z) have a pole of order m at z = a. Then f (z) = (z−a)−mg(z) where g(z)∈ A(B(a; δ)) and g(a) , 0.
Thus limz→a(z− a)mf (z) = g(a) , 0 and
limz→a(z− a)m+1f (z) = lim
z→a(z− a)g(z) = 0.
So
limz→a|z − a|s| f (z)| = limz
→a|z − a|s−m|z − a|m| f (z)| =
0, s > m
∞, s < m by Exercise 7. Therefore, m > 0 is the algebraic order of f at z = a.
Exercise 9. A function f has an essential singularity at z = a iff neither (1.19) nor (1.20) holds for any real number s.
Solution. ⇒: We prove this direction by proving the contrapositive (P → Q ⇐⇒ ¬Q ⇒ ¬P). Assume (1.19) or (1.20) hold for some real number s. Then by Exercise 7, there exists an integer m such that (1.19) holds if s > m and (1.20) holds if s < m. Then by Exercise 8, z = a is either a removable singularity or a pole. Hence z = a is not an essential singularity (by Definition 1.3 p.105).
⇐: Also this direction by is being proved by showing the contrapositive. Assume that f has an isolated singularity at z = a which is not an essential singularity. Then either z = a is a removable singularity or a pole (by Definition 1.3 p.105). Again by Problem 7 and 8, in either case there exists an m such that (1.19) or (1.20) holds. (If z = a is removable, then either m = 0 or m < 0; if z = a is a pole, then m > 0 by Exercise 8).
Exercise 10. Suppose that f has an essential singularity at z = a. Prove the following strengthened version of the Casorati–Weierstrass Theorem. If c∈ C and > 0 are given then for each δ > 0 there is a number α,
|c − α| < , such that f (z) = α has infinitely many solutions in B(a; δ).
Solution. Not available.
Exercise 11. Give the Laurent series development of f (z) = exp1
z
. Can you generalize this result?
Solution. Not available.
Exercise 12. (a) Let λ∈ C and show that exp( 1
2λ z +1 z
!)
= a0+ X∞ n=1
an zn+ 1 zn
!
for 0 <|z| < ∞, where for n ≥ 0
an =1 π
Z π 0
eλ cos tcos nt dt.
(b) Similarly, show
exp( 1 2λ z−1
z
!)
= b0+ X∞
n=1
bn zn+(−1)n zn
!
for 0 <|z| < ∞, where
bn= 1 π
Z π 0
cos(nt− λ sin t) dt.
Solution. Not available.
Exercise 13. Let R > 0 and G ={z : |z| > R}; a function f : G → C has a removable singularity, a pole, or an essential singularity at infinity if f (z−1) has, respectively, a removable singularity, a pole, or an essential singularity at z = 0. If f has a pole at∞ then the order of the pole is the order of the pole of f (z−1) at z = 0.
(a) Prove that an entire function has a removable singularity at infinity iff it is a constant.
(b) Prove that an entire function has a pole at infinity of order m iff it is a polynomial of degree m.
(c) Characterize those rational functions which have a removable singularity at infinity.
(d) Characterize those rational functions which have a pole of order m at infinity.
Solution. Not available.
Exercise 14. Let G ={z : 0 < |z| < 1} and let f : G → C be analytic. Suppose that γ is a closed rectifiable curve in G such that n(γ; a) = 0 for all a in C− G. What isR
γ f ? Why?
Solution. Not available.
Exercise 15. Let f be analytic in G ={z : 0 < |z − a| < r} except that there is a sequence of poles {an} in G with an→ a. Show that for any ω in C there is a sequence {zn} in G with a = lim znand ω = lim f (zn).
Solution. Not available.
Exercise 16. Determine the regions in which the functions f (z) = sin1z−1
and g(z) =R1
0(t− z)−1 dt are analytic. Do they have any isolated singularities? Do they have any singularities that are not isolated?
Solution. Not available.
Exercise 17. Let f be analytic in the region G =ann(a; 0, R). Show that if!
G| f (x + iy)|2dx dy <∞ then f has a removable singularity at z = a. Suppose that p > 0 and!
G| f (x + iy)|pdx dy <∞; what can be said about the nature of the singularity at z = a?
Solution. Not available.