1) The document provides solutions to homework problems from a complex analysis course. It solves problems involving properties of polynomials, including showing two polynomials are equal if they have the same roots, and investigating properties of roots. 2) It also analyzes singularities of complex functions, determining whether singularities are removable, poles, or essential singularities. Functions include rational, trigonometric, exponential and combined functions. 3) The solutions demonstrate techniques for analyzing complex functions at isolated singular points using principles like series expansions and the Casorati-Weierstrass theorem.

1. NCU Math, Spring 2014: Complex Analysis Homework Solution 6
Text Book: An Introduction to Complex Analysis
Problem. 19.1
Sol:
Without loss of generality, we may assume that Pn(z) =
n
k=0 akzk
where ak ∈ R for all k = 0, · · · , n. Since z0
is a root of Pn(z), we have Pn(z0) = 0. By virtue of ak ∈ R for all k = 0, · · · , n and Pn(z0) = 0, we can get that
Pn(z0) =
n
k=0
ak(z0)k
=
n
k=0
ak(z0)k
=
n
k=0
ak · (z0)k
=
n
k=0
ak(z0)k
=
n
k=0
ak(z0)k
= Pn(z0)
= 0.
Thus z0 is also a root of Pn(z).
Problem. 19.2
Sol:
We dene f(z) = P(z) − Q(z). Then f(z) is a polynomial of degree at most n. Since P(z) and Q(z) agree on
n + 1 distinct points, we can get that P(zk) = Q(zk) for k = 1, · · · , n + 1 and zk = zl if k = l. So z1, · · · , zn+1
are distinct roots of f(z). But if P(z) ≡ Q(z), there are at most n distinct roots of f(z) by Corollary 19.1. This
implies that P(z) ≡ Q(z). Thus P(z) = Q(z) for all z ∈ C.
Problem. 19.3
Sol:
It is easy to see that z = 1 is not a root of Pn(z).
1

2. Now, we want to show that for all z ∈ C − {±1} with |z| ≤ 1 is not a root of Pn(z). To see this, we assume
that z0 ∈ C − {±1} with |z0| ≤ 1 is a root of Pn(z). If we can nd a contradiction, we have done what we want to
prove. Since z0 is a root of Pn(z), we have |(z0 − 1)Pn(z0)| = 0. But
(z0 − 1)Pn(z0)
=(z0 − 1)(anzn
0 + · · · + a1z0 + a0)
=anzn+1
0 + (an−1 − an)zn
0 + · · · + (a0 − a1)z0 − a0.
We can get that
0 = |(z0 − 1)Pn(z0)|
≥ |a0| − |anzn+1
0 + (an−1 − an)zn
0 + · · · + (a0 − a1)z0| .
This implies that a0 = |a0| = |anzn+1
0 + (an−1 − an)zn
0 + · · · + (a0 − a1)z0|. Because
a0 = anzn+1
0 + (an−1 − an)zn
0 + · · · + (a0 − a1)z0
≤an|z0|n+1
+ (an−1 − an)|z0|n
+ · · · + (a0 − a1)|z0|
≤an + (an−1 − an) + · · · + (a0 − a1)
=a0,
we have that all inequalities must be equalities. Thus z0 ∈ R from the rst equality and |z0| = 1 from the second
equality. We get that z0 = ±1. This leads a contradiction. Therefore we have showed what we want.
Problem. 30.1
Sol:
(a)
Since
f(z) =
ez2
− 1
z
=
∞
k=0
(z2
)k
k! − 1
z
=
∞
k=1
z2k
k!
z
=
∞
k=1
z2k−1
k!
= z
∞
k=0
z2k
(k + 1)!
→ 0 as z → 0,
we can get that z = 0 is a removable singularity of f(z) by Theorem 29.1. Here, we dene f(0) = 0 such that f is
analytic at z = 0.
2

3. (b)
Since
f(z) =
sin z − z
z2
=
∞
k=0
(−1)k
z2k+1
(2k+1)! − z
z2
=
∞
k=1
(−1)k
z2k+1
(2k+1)!
z2
=
∞
k=0
(−1)k+1
z2k+3
(2k+3)!
z2
=
∞
k=0
(−1)k+1
z2k+1
(2k + 3)!
→ 0 as z → 0,
we can get that z = 0 is a removable singularity of f(z) by Theorem 29.1. Here, we dene f(0) = 0 such that f is
analytic at z = 0.
(c)
Since
f(z) =
1 − 1
2 z2
− cos z
sin z2
=
1 − 1
2 z2
−
∞
k=0
(−1)k
z2k
(2k)!
∞
k=0
(−1)k(z2)2k+1
(2k+1)!
=
−
∞
k=2
(−1)k
z2k
(2k)!
∞
k=0
(−1)kz4k+2
(2k+1)!
=
−
∞
k=0
(−1)k+2
z2k+4
(2k+4)!
z2 ∞
k=0
(−1)kz4k
(2k+1)!
=
−
∞
k=0
(−1)k+2
z2k+2
(2k+4)!
∞
k=0
(−1)kz4k
(2k+1)!
→ 0 as z → 0,
we can get that z = 0 is a removable singularity of f(z) by Theorem 29.1. Here, we dene f(0) = 0 such that f is
analytic at z = 0.
Problem. 30.2
Sol:
3

4. (a)
For convenience, we dene
f(z) =
cos z
z2−π2
/4
when z = ±π
2 ,
− 1
π when z = ±π
2 .
It is easy to see that f is analytic if z = ±π
2 . Since
cos z
z2 − π2
/4
=
∞
k=0
dk(cos z)
dzk
z= π
2
k! (z − π
2 )k
z2 − π2
/4
=
∞
k=1
dk(cos z)
dzk
z= π
2
k! (z − π
2 )k
z2 − π2
/4
=
∞
k=1
dk(cos z)
dzk
z= π
2
k! (z − π
2 )k−1
z + π/2
→
−1
π
as z →
π
2
,
we can get that z = π
2 is a removable singularity of f(z) by Theorem 29.1 and f is analytic at z = π
2 if f(π
2 ) = −1
π .
So f is analytic at z = π
2 . Similarly,
cos z
z2 − π2
/4
=
∞
k=0
dk(cos z)
dzk
z=− π
2
k! (z + π
2 )k
z2 − π2
/4
=
∞
k=1
dk(cos z)
dzk
z=− π
2
k! (z + π
2 )k
z2 − π2
/4
=
∞
k=1
dk(cos z)
dzk
z=− π
2
k! (z + π
2 )k−1
z − π/2
→
−1
π
as z →
π
2
,
we have that z = −π
2 is a removable singularity of f(z) by Theorem 29.1 and f is analytic at z = −π
2 if f(−π
2 ) = −1
π .
Thus f is analytic at z = −π
2 .
Therefore, f is entire.
4

5. (b)
For convenience, we dene
g(z) =
ez
−e−z
2z when z = 0,
1 when z = 0.
It is easy to see that g is analytic if z = 0. Since
ez
− e−z
2z
=
∞
k=0
zk
k! −
∞
k=0
(−z)k
k!
2z
=
∞
k=0
zk
−(−1)k
zk
k!
2z
=
2
∞
k=0
z2k+1
(2k+1)!
2z
=
∞
k=0
z2k
(2k + 1)!
→1 as z → 0,
we can get that z = 0 is a removable singularity of g(z) by Theorem 29.1 and g is analytic at z = 0 if g(0) = 1. So
g is analytic at z = 0.
Therefore, g is entire.
Problem. 30.3
Sol:
(a)
It is easy to see that the isolated singularities are 0 and 1.
Since z3
+1
z2(z−1) =
z3+1/z−1
z2 , z3
+1
z−1 is analytic at z = 1, and limz→0
z3
+1
z−1 = −1 = 0, we have z3
+1
z2(z−1) has a pole of
order 2 at z = 0 by Theorem 29.2.
Because z3
+1
z2(z−1) =
z3+1/z2
z−1 , z3
+1
z2 is analytic at z = 0, and limz→1
z3
+1
z2 = 2 = 0, we have z3
+1
z2(z−1) has a pole of
order 1 at z = 1 by Theorem 29.2.
(b)
It is easy to see that the only isolated singularity is 0.
5

6. Since for z = 0,
z2
e
1/z
= z2
∞
k=0
1
k!
1
zk
= z2
k=0
−∞
1
(−k)!
zk
=
k=0
−∞
1
(−k)!
zk+2
=
k=2
−∞
1
(−k + 2)!
zk
,
we can get that z = 0 is an essential singularity of z2
e1/z
by the denition.
(c)
It is easy to see that the isolated singularities are (2k+1)
2 π for all k ∈ Z.
Since cos z = 0 if z = (2k+1)
2 π for all k ∈ Z and (cos z) = − sin z = (−1)k+1
= 0 if z = (2k+1)
2 π for all k ∈ Z,
we can get that cos z has a zero of order 1 at z = (2k+1)
2 π for all k ∈ Z. We also have z = 0 at z = (2k+1)
2 π for all
k ∈ Z. This implies that z
cos z has a pole of order 1 at z = (2k+1)
2 π for all k ∈ Z by Corollary 29.3.
(d)
It is easy to see that the only isolated singularity is 0.
Since z2
has a zero of order 2 at z = 0 and sin 3z =
∞
k=0
(−1)k
(3z)2k+1
(2k+1)! has a zero of order 1 at z = 0 by Theorem
26.1, we have sin 3z
z2 has a pole of order 1 at z = 0 by Corollary 29.3.
(e)
It is easy to see that the isolated singularities are 0 and ±i.
Since cos 1
z =
∞
k=0
(−1)k
z−2k
(2k)! =
k=0
−∞
(−1)−k
z2k
(−2k)! , we can get that cos 1
z has an essential singularity at z = 0. So
by Theorem 30.1 (Casorati-Weierstrass Theorem), we have
1. there exists a sequence {αn} such that αn → 0 and limn→∞ | cos 1
αn
| = ∞.
2. for any complex number ω, there exists a sequence βn (depending on ω) such that βn → 0 and limn→∞ cos 1
βn
=
−ω
sin 1 .
Because sin(z−1)
z2+1 → − sin 1 as z → 0, we can get that limn→∞ | cos 1
αn
sin(αn−1)
α2
n+1 | = ∞ and limn→∞ cos 1
βn
sin(βn−1)
β2
n+1 =
ω. This implies that cos 1
z
sin(z−1)
z2+1 has an essential singularity at z = 0 by Theorem 30.1 (Casorati-Weierstrass
Theorem).
Since cos 1
z
sin(z−1)
z2+1 =
cos 1
z
sin(z−1)/z+i
z−i , cos 1
z sin(z−1)/z+i is analytic if z = −i and z = 0, and cos 1
z sin(z−1)/z+i|z=i =
cos(−i) sin(i−1)
2i = 0, we can get that cos 1
z
sin(z−1)
z2+1 has a pole of order 1 at z = i by Theorem 29.2.
Similarly, from cos 1
z sin(z−1)/z−i is analytic if z = i and z = 0, cos 1
z sin(z−1)/z−i|z=−i = cos i sin(−i−1)
−2i = 0, and
Theorem 29.2, we have cos 1
z
sin(z−1)
z2+1 =
cos 1
z
sin(z−1)/z−i
z+i has a pole of order 1 at z = −i.
6

7. (f)
It is easy to see that the isolated singularities are 1
2 , ±i, and k ∈ Z.
Since (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
=
(z2−1) cos πz/(z+i)2(z+2)(2z−1) sin2 πz
(z−i)2 , (z2
−1) cos πz/(z+i)2
(z+2)(2z−1) sin2
πz is analytic at
z = i, and (z2
−1) cos πz/(z+i)2
(z+2)(2z−1) sin2
πz
z=i
= −2 cos(iπ)
−4(i+2)(2i−1) sin2(iπ)
= 0, we can get that (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
has a pole of order 2 at z = i by Theorem 29.2.
Similarly, from (z2
−1) cos πz/(z−i)2
(z+2)(2z−1) sin2
πz is analytic at z = −i, (z2
−1) cos πz/(z−i)2
(z+2)(2z−1) sin2
πz
z=−i
=
−2 cos(−iπ)
−4(−i+2)(−2i−1) sin2(−iπ)
= 0, and Theorem 29.2, we have (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
=
(z2−1) cos πz/(z−i)2(z+2)(2z−1) sin2 πz
(z+i)2
has a pole of order 2 at z = −i.
Because
(z −
1
2
)
(z2
− 1) cos πz
(z + 2)(2z − 1)(z2 + 1)2 sin2
πz
=
(z2
− 1) cos πz
2(z + 2)(z2 + 1)2 sin2
πz
→
(1
4 − 1) · 0
2(1
2 + 2)(1
4 + 1)2 sin2
(π
2 )
= 0 as z →
1
2
,
we can get that z = 1
2 is a removable singularity of (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
by Theorem 29.1.
From Example 26.1, we have sin kπ has a zero of order 1 at z = k ∈ Z. By virtue of Corollary 26.1, we can get
that sin2
πz has a zero of order 2 at z = k ∈ Z.
Since z + 2 has a zero of order 1 at z = −2 and sin2
πz has a zero of order 2 at z = k ∈ Z, we can get
that (z + 2) sin2
πz has a zero of order 3 at z = −2. We also have that (z2
−1) cos πz
(2z−1)(z2+1)2 is analytic at z = −2 and
(z2
−1) cos πz
(2z−1)(z2+1)2
z=−2
= (4−1) cos(−2π)
(−4−1)(4+1)2 = 3
−125 = 0. This implies that (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
=
(z2−1) cos πz/(2z−1)(z2+1)2
(z+2) sin2 πz
has a pole of order 3 at z = −2 by Corollary 29.3.
From (z+1) cos πz
(z+2)(2z−1)(z2+1)2
z=1
= (1+1) cos π
(1+2)(2−1)(1+1)2 = −2
12 = 0 and (z+1) cos πz
(z+2)(2z−1)(z2+1)2 is analytic at z = 1, we can get
that (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 = (z − 1) (z+1) cos πz
(z+2)(2z−1)(z2+1)2 has a zero of order 1 at z = 1. By above, sin2
πz has a zero of
order 2 at z = k ∈ Z, and Corollary 29.3, we have z = 1 is a pole of order 1 of (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
.
Similarly, we can get that (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 = (z + 1) (z−1) cos πz
(z+2)(2z−1)(z2+1)2 has a zero of order 1 at z = −1. So
z = −1 is a pole of order 1 of (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
by virtue of above, sin2
πz has a zero of order 2 at z = k ∈ Z,
and Corollary 29.3.
Since (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 is analytic at z = k ∈ Z−{±1, −2}, (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2
z=k
= 0 for all z ∈ Z−{±1, −2},
sin2
πz has a zero of order 2 at z = k ∈ Z, and Corollary 29.3, we have (z2
−1) cos πz
(z+2)(2z−1)(z2+1)2 sin2 πz
has a pole of order
2 at z ∈ Z − {±1, −2}.
Problem. 30.4
Sol:
7

8. Since f(z) has a pole of order m at z0, we can nd R 0 such that for z ∈ C with 0 |z − z0| R,
f(z) =
∞
k=−m ak(z − z0)k
where a−m ∈ C − {0}. So we can get that for z ∈ C with 0 |z − z0| R,
f (z) =
∞
k=−m
kak(z − z0)k−1
=
∞
k=−m−1
(k + 1)ak+1(z − z0)k
.
Then we have f (z) has a pole of order m + 1 at z0 from the denition of singularity.
Problem. 31.1
Sol:
(a)
From Theorem 31.1, we have
R
z3
+ 1
z2(z − 1)
, 0 = lim
z→0
1
1!
d
dz
z2
·
z3
+ 1
z2(z − 1)
= lim
z→0
d
dz
z3
+ 1
z − 1
= lim
z→0
3z2
(z − 1) − (z3
+ 1) · 1
(z − 1)2
= −1.
By virtue of Corollary 31.1, we can get that
R
z3
+ 1
z2(z − 1)
, 1 = lim
z→1
(z − 1) ·
z3
+ 1
z2(z − 1)
= lim
z→1
z3
+ 1
z2
= 2.
(b)
Since for z = 0, z2
e1/z
=
k=2
−∞
1
(−k+2)! zk
, we can get that R z2
e1/z
, 0 = 1
3! = 1
6 by the denition.
(c)
By virtue of Corollary 31.1, we can get that for all k ∈ Z, R z
cos z , (2k+1)
2 π = limz→
(2k+1)
2 π
(z − (2k+1)
2 π) · z
cos z .
Since limz→
(2k+1)
2 π
cos z
z−
(2k+1)
2 π
= − sin z|z=
(2k+1)
2 π
= (−1)k+1
and limz→
(2k+1)
2 π
z = (2k+1)
2 π for all k ∈ Z, we have for
8

9. all k ∈ Z,
R
z
cos z
,
(2k + 1)
2
π = lim
z→
(2k+1)
2 π
(z −
(2k + 1)
2
π) ·
z
cos z
=
(−1)k+1
(2k + 1)
2
π.
(d)
By virtue of Corollary 31.1, we can get that
R
sin 3z
z2
, 0 = lim
z→0
z ·
sin 3z
z2
= lim
z→0
sin 3z
z
= 3 cos 3z|z=0
= 3.
Problem. 31.2
Sol:
(a)
From Theorem 29.2, z2n
(z+1)n has a pole of order n at z = −1. So by Theorem 31.1, we have
R
z2n
(z + 1)n
, −1 = lim
z→−1
1
(n − 1)!
dn−1
dzn−1
(z + 1)n
·
z2n
(z + 1)n
= lim
z→−1
1
(n − 1)!
dn−1
dzn−1
z2n
= lim
z→−1
2n · (2n − 1) · · · · · (2n − n + 2)
(n − 1)!
z2n−n+1
= C2n
n−1(−1)n+1
.
(b)
9

10. From Problem 30.3 (e) and Corollary 31.1, we have
R cos
1
z
sin(z − 1)
z2 + 1
, i = lim
z→i
(z − i) · cos
1
z
sin(z − 1)
z2 + 1
= lim
z→i
cos
1
z
sin(z − 1)
z + i
=
cos(−i) sin(i − 1)
2i
=
ei(−i)
+e−i(−i)
2 · ei(i−1)
−e−i(i−1)
2i
2i
=
(e + e−1
) · (e−1−i
− e1+i
)
−8
=
cosh 1 · sinh(1 + i)
2
.
(c)
From Theorem 29.2, cos z
z2(z−π)3 has a pole of order 2 at z = 0. So by Theorem 31.1, we have
R
cos z
z2(z − π)3
, 0 = lim
z→0
1
1!
d
dz
z2
·
cos z
z2(z − π)3
= lim
z→0
d
dz
cos z
(z − π)3
= lim
z→0
−(z − π)3
sin z − 3(z − π)2
cos z
(z − π)6
= −
3
π4
.
(d)
From Theorem 29.2, cos z
z2(z−π)3 has a pole of order 3 at z = π. So by Theorem 31.1, we have
R
cos z
z2(z − π)3
, π = lim
z→π
1
2!
d2
dz2
(z − π)3
·
cos z
z2(z − π)3
=
1
2
lim
z→π
d2
dz2
cos z
z2
=
1
2
lim
z→π
d
dz
−z2
sin z − 2z cos z
z4
=
1
2
lim
z→π
z4
(−2z sin z − z2
cos z − 2 cos z + 2z sin z) − 4z3
(−z2
sin z − 2z cos z)
z8
=
1
2
π4
(π2
+ 2) − 4π3
(2π)
π8
=
π2
− 6
2π4
.
10

11. (e)
Since
z3
cos
1
z − 2
= (z − 2)3
+ 8(z − 2)2
+ 12(z − 2) + 8 ·
k=0
−∞
(−1)−k
(z − 2)2k
(−2k)!
,
we can get that
R z3
cos
1
z − 2
, 2 = 1 ·
(−1)2
4!
+ 8 · 0 + 12 ·
(−1)1
2!
+ 8 · 0
= −
143
24
by virtue of the denition.
(f)
Since for 0 |z| 1,
e
1
z
z2 + 1
=
1
1 − (−z2)
· e
1
z
=
∞
k=0
(−1)k
z2k
·
∞
k=0
z−k
k!
=
∞
k=0
(−1)k
z2k
·
k=0
−∞
zk
(−k)!
,
we can get that
R
e
1
z
z2 + 1
, 0 =
∞
k=0
(−1)k 1
(2k + 1)!
= sin 1.
by virtue of the denition.
Problem. 31.4
Sol:
(a)
11

12. It is easy to see that z2
(z−1)2(z+2) has a pole of order 2 at z = 1 and a pole of order of 1 at z = −2. So
R
z2
(z − 1)2(z + 2)
, 1 = lim
z→1
1
1!
d
dz
(z − 1)2
·
z2
(z − 1)2(z + 2)
= lim
z→1
d
dz
z2
(z + 2)
= lim
z→1
2z(z + 2) − z2
(z + 2)2
=
5
9
and
R
z2
(z − 1)2(z + 2)
, −2 = lim
z→−2
(z + 2) ·
z2
(z − 1)2(z + 2)
= lim
z→−2
z2
(z − 1)2
=
4
9
.
From Theorem 31.2, we have
ˆ
γ
z2
(z − 1)2(z + 2)
dz = 2πi R
z2
(z − 1)2(z + 2)
, 1 + R
z2
(z − 1)2(z + 2)
, −2
= 2πi.
(b)
From Theorem 31.2 and z3
(z−2)(z−1−i) is analytic for all |z| 1.5 with z = 1 + i, we have
ˆ
γ
z3
(z − 2)(z − 1 − i)
dz = 2πi · R
z3
(z − 2)(z − 1 − i)
, 1 + i .
It is easy to see that z3
(z−2)(z−1−i) has a pole of order 1 at z = 1 + i. So
R
z3
(z − 2)(z − 1 − i)
, 1 + i = lim
z→1+i
(z − 1 − i) ·
z3
(z − 2)(z − 1 − i)
= lim
z→1+i
z3
(z − 2)
=
(1 + i)3
−1 + i
= 2.
This implies that
´
γ
z3
(z−2)(z−1−i) dz = 4πi.
(c)
12

13. From Theorem 31.2 and 1
z4−1 is analytic inside γ with z = 1, z = i, and z = −i, we have
ˆ
γ
1
z4 − 1
dz = 2πi · R
1
z4 − 1
, 1 + R
1
z4 − 1
, i + R
1
z4 − 1
, −i .
It is easy to see that 1
z4−1 has a pole of order 1 at z = 1, ±i. So
R
1
z4 − 1
, 1 = lim
z→1
(z − 1) ·
1
z4 − 1
= lim
z→1
1
(z + 1)(z + i)(z − i)
=
1
2(1 + i)(1 − i)
=
1
4
.
Similarly, we can get that R 1
z4−1 , i = 1
(i+1)(i−1)2i = i
4 and R 1
z4−1 , −i = 1
(−i+1)(−i−1)(−2i) = − i
4 . This implies
that
´
γ
1
z4−1 dz = πi
2 .
(d)
From Theorem 31.2 and z
(z2−1)2(z2+1) is analytic inside γ with z = 1, z = i, and z = −i, we have
ˆ
γ
z
(z2 − 1)2(z2 + 1)
dz = 2πi · R
z
(z2 − 1)2(z2 + 1)
, 1 + R
z
(z2 − 1)2(z2 + 1)
, i + R
z
(z2 − 1)2(z2 + 1)
, −i .
It is easy to see that z
(z2−1)2(z2+1) has a pole of order 1 at z = ±i and a pole of order 2 at z = 1. So
R
z
(z2 − 1)2(z2 + 1)
, 1 = lim
z→1
1
1!
d
dz
(z − 1)2
·
z
(z2 − 1)2(z2 + 1)
= lim
z→1
d
dz
z
(z + 1)2(z2 + 1)
= lim
z→1
(z + 1)2
(z2
+ 1) − z(2(z + 1)(z2
+ 1) + 2z(z + 1)2
)
(z + 1)4(z2 + 1)2
=
4 · 2 − (2 · 2 · 2 + 2 · 4)
16 · 4
= −
1
8
.
We can also get that
R
z
(z2 − 1)2(z2 + 1)
, i = lim
z→i
(z − i) ·
z
(z2 − 1)2(z2 + 1)
= lim
z→i
z
(z2 − 1)2(z + i)
=
i
4 · 2i
=
1
8
.
13

14. Similarly, R z
(z2−1)2(z2+1) , −i = −i
4·(−2i) = 1
8 . This implies that
´
γ
1
z4−1 dz = πi
4 .
(e)
Since for |z| 0, we have
cos(
1
z2
)e
1
z =
∞
k=0
(−1)k
(2k)!
1
z4k
·
∞
k=0
1
k!
1
zk
=
k=0
−∞
(−1)−k
(−2k)!
z−4k
·
k=0
−∞
1
(−k)!
z−k
.
Then we can get that
R cos(
1
z2
)e
1
z , 0 =
(−1)0
0!
·
1
1!
= 1
by the denition. So by Theorem 31.2 and cos( 1
z2 )e
1
z is analytic at z = 0, we have
ˆ
γ
cos(
1
z2
)e
1
z dz = 2πi · R cos(
1
z2
)e
1
z , 0
= 2πi.
(f)
Because tan z = sin z
cos z and cos z has a zero of order 1 at z = (2k+1)
2 π for all k ∈ Z, we can get that for all k ∈ Z,
R tan z,
(2k + 1)
2
π = lim
z→
(2k+1)
2 π
(z −
(2k + 1)
2
π)
sin z
cos z
= lim
z→
(2k+1)
2 π
sin z
cos z
(z−
(2k+1)
2 π)
.
Since limz→
(2k+1)
2 π
sin z = (−1)k
and limz→
(2k+1)
2 π
cos z
(z−
(2k+1)
2 π)
= (− sin z)|z=
(2k+1)
2 π
= (−1)k+1
, we have R tan z, (2k+1)
2 π =
−1 for all k ∈ Z.
From Theorem 31.2, we have
ˆ
γ
tan zdz = 2πi · R tan z,
−π
2
+ R tan z,
π
2
= −4πi.
Problem. 31.5
Sol:
14

15. Let γ be the circle |z + 1| = 2 that traversed once in the counterclockwise direction. So
´
γ
dz
(z2+z)2 =
−2
´
γ
dz
(z2+z)2 .
It is easy to see that 1
(z2+z)2 = 1
z2(z+1)2 has a pole of order 2 at z = 0 and z = −1. Then
R
1
(z2 + z)2
, 0 = lim
z→0
1
1!
d
dz
z2
·
1
(z2 + z)2
= lim
z→0
d
dz
1
(z + 1)2
= lim
z→0
−2
(z + 1)3
= −2
and
R
1
(z2 + z)2
, −1 = lim
z→−1
1
1!
d
dz
(z + 1)2
·
1
(z2 + z)2
= lim
z→−1
d
dz
1
z2
= lim
z→−1
−2
z3
= 2.
By virtue of Theorem 31.2, we can get that
ˆ
γ
1
(z2 + z)2
dz = 2πi · R
1
(z2 + z)2
, 0 + R
1
(z2 + z)2
, −1
= 0.
This implies that
´
γ
dz
(z2+z)2 = −2
´
γ
dz
(z2+z)2 = 0.
Problem. 31.6
Sol:
Since f has an isolated singularity at 0, we can nd R 0 and ak ∈ C for all k ∈ Z such that f(z) =
∞
k=−∞ akzk
for all 0 |z| R. Because f is an even function, we have a2j+1 = 0 for all j ∈ Z by Problen 25.7. This implies
that R[f, 0] = a−1 = 0 from the denition.
15