Math 311 - Spring 2014
Solutions to Assignment # 11
Completion Date: Tuesday June 10, 2014
Question 1. [p 239, #1]
Find the residue at z = 0 of the function
1 1 z sin z cot z sinh z
(a) 2
; (b) z cos ; (c) ; (d) ; (e) .
z+z z z z4 z 4 (1 z 2)
Ans: (a) 1; (b) 1/2; (c) 0; (d) 1/45; (e) 7/6.
Solution:
(a) For 0 < |z| < 1, we have
1 1 1 1 1
2
= = 1 z + z2 z3 + = 1 + z z2 + ,
z+z z 1+z z z
so the residue at z = 0 is 1.
(b) For z 6= 0, we have
1 1 1 1 1 1 1 1 1
z cos = z 1 2 + 4 + = z + 3 + ,
z 2! z 4! z 2! z 4! z
1
so the residue at z = 0 is .
2!
(c) For z 6= 0, we have
z sin z 1 z3 z5 z2 z4
= z z + + = + ,
z z 3! 5! 3! 5!
and there are no negative powers of z, so the residue at z = 0 is 0.
(d) For 0 < |z| < , we have
z2 z4 z6
cot z cos z 1 + +
4
= 4 = 2! 3 4! 5 6! .
z z sin z 4
z z
z z + +
3! 5!
Now let
z2 z4 z6
w= + + ,
3! 5! 7!
then for z 6= 0 such that |w| < 1, we have
z2 z4 z6
cot z 1 + + 1 z2 z4
= 2! 4! 6! = 1 + + 1 + w + w 2
+ w 3
+ .
z4 z 5 (1 w) z5 2! 4!
Therefore, for 0 < |z| < , we have
cot z 1 z2 z4 z2 1 1
4
= 5 1 + + 1+ + z4 + ,
z z 2! 4! 3! (3!)2 5!
that is,
cot z 1 1 1 1 1 1 1 1 1
= 5 + + +
z4 z 2! 3! z3 (3!)2 5! 4! 2!3! z
for 0 < |z| < , and
cot z 1 1 1 1 1
Res = + = .
z=0 z4 (3!) 2 5! 4! 2!3! 45
(e) For 0 < |z| < 1, we have
sinh z 1 z2 z4
4 2
= 3
1 + + + 1 + z2 + z4 + ,
z (1 z ) z 3! 5!
so that
sinh z 1 1 2
= 3 1+ 1+ z + + higher order terms ,
z 4 (1 z 2 ) z 3!
that is,
sinh z 1 7 1
= 3 + +
z 4 (1 z 2 ) z 6 z
for 0 < |z| < 1, and
sinh z 7
Res = .
z=0 z (1 z 2 )
4 6
Question 2. [p 239, #2 (a)]
Use Cauchys residue theorem (Sec. 70) to evaluate the integral of
exp(z)
z2
around the circle |z| = 3 in the positive sense.
Ans: 2i.
ez
Solution: The function has an isolated singularity at z = 0 which is inside the circle |z| = 3, and since
z2
ez 1 z2 z3 1 1 1 z
= 1 z + + = 2 + +
z2 z2 2! 3! z z 2! 3!
for 0 < |z| < , then
ez
Res = 1.
z=0 z2
ez
Now, since is analytic inside and on |z| = 3, except at z = 0, then
z2
I z
ez e
2
dz = 2i Res = 2i.
|z|=3 z z=0 z2
Question 3. [p 239, #2 (c)]
Use Cauchys residue theorem (Sec. 70) to evaluate the integral of
2 1
z exp
z
around the circle |z| = 3 in the positive sense.
Ans: i/3.
1
Solution: The function z 2 exp has an isolated (essential) singularity at z = 0 which is inside the circle
z
|z| = 3, and since
1 1 1 1 1 1 1 1
z 2 exp = z2 1 + + 2 + 3 + 4 +
z z 2! z 3! z 4! z
for 0 < |z| < , then
1 1
Res z 2 exp = .
z=0 z 3!
2 1
Now, since z exp is analytic inside and on |z| = 3, except at z = 0, then
z
I
1 1 1 i
z 2 exp dz = 2i Res z 2 exp = 2i = .
|z|=3 z z=0 z 3! 3
Question 4. [p 239, #2 (d)]
Use Cauchys residue theorem (Sec. 70) to evaluate the integral of
z+1
z 2 2z
around the circle |z| = 3 in the positive sense.
Ans: 2i.
z+1 z+1
Solution: The function 2
= has isolated singularities at z = 0 and z = 2, both of which lie
z 2z z(z 2)
inside the circle |z| = 3, so that
I
z+1 z+1 z+1
2
dz = 2i Res + Res
|z|=3 z 2z z=0 z 2 2z z=2 z 2 2z
Now, since
z+1 1 1 3 1 1 1
= + = ,
z(z 2) z 2 z(z 2) 2 z2 2 z
1
and since is analytic at z = 0, then
z2
z+1 1
Res = ,
z=0 z 2 2z 2
1
while since is analytic at z = 2, then
z
z+1 3
Res = ,
z=2 z 2 2z 2
and I
z+1 1 3
dz = 2i + = 2i.
|z|=3 z 2 2z 2 2
Question 5. [p 243, #1]
In each case, write the principal part of the function at its isolated singular point and determine whether
that point is a pole, a removable singular point, or an essential singular point:
1 z2 sin z cos z 1
(a) z exp ; (b) ; (c) ; (d) ; (e) .
z 1+z z z (2 z)3
Soution:
(a) For 0 < |z| < , we have
1 1 1 1 1 1 1
z exp = z +1+ + 2 ++ n1 +
z 2!
| z 3! z {z n! z }
principal part
and the isolated singular point z = 0 is an essential singular point since the principal part has
infinitely many nonzero terms.
(b) For z 6= 1, we have
z2 1
=1+z2+
z+1 1+z
| {z }
principal part
the isolated singular point z = 1 is a simple pole, that is, a pole of order 1.
(c) For 0 < |z| < , we have
sin z z2 z4 z6
=1 + + ,
z 3! 5! 7!
and the isolated singular point z = 0 is a removable singular point, since there are no nonzero
terms in the principal part.
(d) For 0 < |z| < , we have
cos z 1 z z3 z5
= + +
z z
|{z} 2! 4! 6!
principal part
and the isolated singular point z = 0 is a simple pole, that is, a pole of order 1.
(e) For z 6= 2, we have
1 1
=
(2 z)3 (z 2)3
| {z }
principal part
and the isolated singular point z = 2 is a pole of order 3.
Question 6. [p 243, #2]
Show that the singular point of each of the following functions is a pole. Determine the order m of that pole
and the corresponding residue B.
1 cosh z 1 exp(2z) exp(2z)
(a) ; (b) ; (c) .
z3 z4 (z 1)2
Ans: (a) m = 1, B = 1/2; (b) m = 3, B = 4/3; (c) m = 2, B = 2e2 .
Solution:
(a) For z 6= 0 we have
1 cosh z 1 z2 z4 1 1 z z3
3
= 3 1 1+ + + = ,
z z 2! 4! 2! z 4! 6!
1
and the isolated singularity z = 0 is a simple pole, that is, a pole of order 1, with residue B = .
2!
(b) For z 6= 0, we have
1 e2z 1 2z 22 z 2 23 z 3 24 z 4 1 1 4 1 2 4
= 1 1 + + + + + = 2 3 2 2 z ,
z4 z4 1! 2! 3! 4! z z 3 z 3 15
4
and the isolated singular point z = 0 is a pole of order 3, with residue B = .
3
(c) For z 6= 1, we have
e2z e2 2(z1) e2 22 (z 1)2 23 (z 1)3
= e = 1 + 2(z 1) + + +
(z 1)2 (z 1)2 (z 1)2 2! 3!
and the isolated singular point z = 1 is a pole of order 2 with residue B = 2e2 .
Question 7. [p 248, #1]
In each case, show that any singular point of the function is a pole. Determine the order m of each pole,
and find the corresponding residue B.
3
z2 + 2 z exp(z)
(a) ; (b) ; (c) .
z1 2z + 1 z 2 + 2
Ans: (a) m = 1, B = 3; (b) m = 3, B = 3/16; (c) m = 1, B = i/2.
Solution:
(a) Note that
z2 + 2 (z)
=
z1 z1
z2 + 2
where (z) = z 2 + 2 is analytic at z = 1, and (1) = 3 6= 0, so that has a simple pole at z = 1
z1
so that m = 1 and
z2 + 2
B = Res = (1) = 3.
z=1 z1
(b) Note that
3
z z3 1 (z)
= 3 = 3
2z + 1 8 1 1
z+ z+
2 2
3
z3 1 1 1 z
where (z) = is analytic at z = and = 6 0, so that
= has a pole of
8 2 2 64 2z + 1
1
order 3 at z = , so that m = 3 and
2
z3 00 (1/2) 1 1 3
B = Res 3
= = 3 2 (1/2) = .
z=1/2 (2z + 1) 2! 2! 8 16
(c) Note that
ez 1 (z)
=
z 2 + 2 z i
where
ez
1 (z) =
z + i
ei ez
is analytic at z = i, and 1 (i) = 6= 0, so that 2 has a simple pole at z = i and
2i z + 2
ez ei 1 i
B1 = Res = = = .
z=i z 2 + 2 2i 2i 2
Also,
ez 2 (z)
=
z 2 + 2 z + i
where
ez
2 (z) =
z i
ei ez
is analytic at z = i, and 2 (i) = 6= 0, so that 2 has a simple pole at z = i and
2i z + 2
ez ei 1 i
B2 = Res 2 2
= = = .
z=i z + 2i 2i 2
Question 8. [p 248, #3]
Find the value of the integral Z
3z 3 + 2
dz,
C (z 1)(z 2 + 9)
taken counterclockwise around the circle
(a) |z 2| = 2; (b) |z| = 4.
Ans: (a) i; (b) 6i.
Solution:
(a) Note that
3z 3 + 2 (z)
f (z) = 2
=
(z 1)(z + 9) z1
where
3z 3 + 2
(z) =
z2 + 9
5 1 1
is analytic at z = 1 and (1) = = 6= 0, so that f has simple pole at z = 1 with residue .
10 2 2
Since f is analytic inside and on the circle C : |z 2| = 2, except at z = 1, which is inside C, then
I
3z 3 + 2 3z 3 + 2 1
2
dz = 2i Res 2
= 2i = i.
|z2|=2 (z 1)(z + 9) z=1 (z 1)(z + 9) 2
(b) Now the singular points of
3z 3 + 2
f (z) =
(z 1)(z 2 + 9)
are z = 1, z = 3i, z = 3i and are all inside the circle C : |z| = 4, and (check this!) since f has
simple poles at z = 1, 3i, and since
1
Res (f (z)) = lim (z 1)f (z) =
z=1 z1 2
2 81i
Res (f (z)) = lim (z 3i)f (z) =
z=3i z3i 6i(3i 1)
2 + 81i
Res (f (z)) = lim (z + 3i)f (z) =
z=3i z3i 6i(3i + 1)
then I
3z 3 + 2 1 2 81i 2 + 81i
dz = 2i + + = 2i 3 = 6i.
|z|=4 (z 1)(z 2 + 9) 2 6i(3i 1) 6i(3i + 1)
Question 9. [p 255, #2]
Show that
z sinh z i
(a) Res = ;
z=i z 2 sinh z
exp(zt) exp(zt)
(b) Res + Res = 2 cos t.
z=i sinh z z=i sinh z
Solution:
(a) Note that
p(z)
f (z) =
q(z)
where
p(z) = z sinh z and q(z) = z 2 sinh z
are analytic at z = i.
Now, the zeros of sinh z are at z = ni for n = 0, 1, 2, . . . , and
p(i) = i sinh(i) = i 6= 0,
while
q(i) = (i)2 sinh(i) = 0,
but
q 0 (i) = 2i sinh(i) + (i)2 cosh(i) = 2 cos = 2 6= 0,
so that q(z) has a simple zero at z = i.
p(z)
Therefore f (z) = has a simple pole at z = i, and
q(z)
z sinh z p(i) i i
B = Res 2
= 0 = 2 = .
z=i z sinh z q (i)
(b) Note that the function
ezt
f (z) =
sinh z
is analytic except where sinh z = 0, that is, except at z = ni, n = 0, 1, 2, . . . .
Now,
ezt
lim zf (z) = lim = 1,
z0 z0 sinh z
z
and f has a simple pole at the isolated singularity z = 0 with Res(f (z)) = 1.
z=0
Also, since sinh(z ni) = (1)n sinh z, then
(1)n ezt
lim (z ni)f (z) = lim = (1)n enit ,
zni zni sinh(z ni)
z ni
and f has a simple pole at the isolated singularity z = ni for each n = 1, 2, . . . with residue
(1)n enit .
Therefore,
exp(zt) exp(zt)
Res + Res = eit eit = 2 cos t
z=i sinh z z=i sinh z
from Eulers formula.
