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Math 421 – Practice Final Solutions
May 12, 2009
- Find the Taylor series or Laurent series of f (z) = (^) z −^2 1 in the following domains.
(a) D 1 = {z | |z| < 1 }.
Solution: Since |z| < 1, f (z) = 2 z − 1
= −^2
1 − z
∑^ ∞
n=
zn^ =
∑^ ∞
n=
(−2)zn.
(b) D 2 = {z | 1 < |z| < ∞}. Solution: Note that 1 < |z| < ∞ ⇒ | 1 z
| < 1, so we have
f (z) =
z − 1 =^
z(1 − (^1) z )
z
∑^ ∞
n=
z )
n (^) =
∑^ ∞
n=
zn+^.
(c) D 5 = {z | 1 < |z − 2 | < ∞}. Solution: Note that 1 < |z − 2 | < ∞ ⇒ | 1 z − 2
| < 1, so we have
f (z) =
z − 1 =^
(z − 2) + 1 =^
(z − 2)(1 − (^) z−−^12 )
z − 2
∑^ ∞
n=
z − 2 )
n (^) =
∑^ ∞
n=
2(−1)n (z − 2)n+^.
- We know f (z) = sin(sin z) − sin z has a zero at z 0 = 0. Find its order.
Solution: Since sin z =
∑^ ∞
n=
(−1)n (2n + 1)! z
2 n+1 (^) = z − z^3 6 +^
z^5 120 − · · ·^ for all^ z, we have
sin(sin z) = sin z − (sin^ z)
3 6 +
(sin z)^5 120 − · · ·
= (z − z
3 6
5 120
(z − z 63 + 120 z^5 − · · · )^3 6
(z − z 63 + 120 z^5 − · · · )^5 120
= z − z
3 6
− z
3 6
3 3
It follows that
f (z) = sin(sin z) − sin z = (z − z
3 3
3 6
3 6
Thus f has a zero of order 3 at z 0 = 0.
- Find the residues of f at all its isolated singular points.
(a) f (z) =
z^2 (z^2 + 1)^2. Solution: To find singular points, set (z^2 + 1)^2 = 0 and solve it to obtain z = ±i.
Resz=i
z^2 (z^2 + 1)^2 = Resz=i
z^2 (z+i)^2 (z − i)^2 =^ φ
′(i), where φ(z) = z^2 (z + i)^2.
Since φ′(z) =^2 z(z^ +^ i)
(^2) − z (^2) (2(z + i)) (z + i)^4 , we have
Resz=i
z^2 (z^2 + 1)^2 =^ φ
′(i) =^2 i(i^ +^ i)^2 −^ i^2 (2(i^ +^ i)) (i + i)^4 =^ −^
i
Similarly, Resz=−i^ z
2 (z^2 + 1)^2
= Resz=−i
z^2 (z−i)^2 (z + i)^2
= ψ′(−i), where ψ(z) = z
2 (z − i)^2
Since ψ′(z) =
2 z(z − i)^2 − z^2 (2(z − i)) (z − i)^4 , we have
Resz=−i^ z
2 (z^2 + 1)^2
= ψ′(−i) = 2(−i)(−i^ −^ i)
(^2) − (−i) (^2) (2(−i − i)) (−i − i)^4
= i 4
(b) f (z) = z
n (z + 3)n^
, where n is a positive integer. Solution: The only singular point of f is z = −3. To find the residue of f at z = −3, denote φ(z) = zn. Then
Resz=− 3
zn (z + 3)n^ =^
φ(n−1)(−3) (n − 1)!. Since φ(n−1)(z) = n(n − 1) · · · 2 · z = n!z, we have Resz=− 3 z
n (z + 3)n^
= φ
(n−1)(−3) (n − 1)!
= n!(−3) (n − 1)!
= − 3 n.
(c) f (z) = cotz z.
Solution: Note that cot^ z z
= cos^ z z sin z
. To find its singular points, set z sin z = 0, and solve it to obtain z = nπ, n ∈ Z. Denote by p(z) = cos z and q(z) = z sin z. Then q′(z) = sin z + z cos z. If n 6 = 0, then q(nπ) = 0 and q′(nπ) = sin(nπ) + nπ cos(nπ) = nπ cos(nπ) 6 = 0. Thus
Resz=nπ
cos z z sin z =^
p(nπ) q′(nπ) =^
cos(nπ) nπ cos(nπ) =^
nπ.
To find the residue of f at 0, note that
z sin z = z(z − z
3 6
2 6
Thus
Resz=0z^ cos sin^ z z = Resz=0^ cos^ z z^2 (1 − z 62 + · · · )
= Resz=
cos z 1 − z 62 +··· z^2 =^ φ
where φ(z) =
cos z 1 − z 62 + · · ·
(c)
|z|=
e(^1 z^ )^ sin(^1 z
) dz.
Solution: The only singular point of e(^
(^1) z ) sin(^1 z
) is 0, which is inside the circle |z| = 1.
To find the residue of e(^
(^1) z ) sin(^1 z
) at 0, we need to compute its Laurent series in a deleted neighborhood of 0. Note that
e(^1 z^ )^ = 1 +^1 z
2!z^2
and sin (^1 z
) =^1
z
3!z^3
So we have
e(^
(^1) z ) sin(
z ) = (1 +
z +^
2!z^2 +^ · · ·^ )(
z −^
3!z^3 +^ · · ·^ ) =
z +
z^2 +^ · · ·^. It follows that Resz=0e(^
(^1) z ) sin(
z ) = 1. Therefore (^) ∮
|z|=
e(^
(^1) z ) sin(
z )^ dz^ = 2πi.
- Evaluate the following integrals.
(a)
∫ (^2) π
0
sin^2 θ 5 + 4 cos θ dθ. Solution: Make the change of variables so the integral will be around the unit circle. z = eiθ^ ⇒ cos θ =
z + (^1) z 2 ,^ sin^ θ^ =^
z − (^1) z 2 i ,^ and^ dθ^ =^
dz iz. So we have ∫ (^2) π
0
sin^2 θ 5 + 4 cos θ dθ^ =
|z|=
( (^21) i (z − (^1) z ))^2 5 + 4( 12 (z + (^1) z ))
dz iz =^
i 4
|z|=
z^4 − 2 z^2 + 1 z^2 (2z^2 + 5z + 2) dz. Set z^2 (2z^2 + 5z + 2) = 0 and solve it to obtain z = 0, − 2 , − 12. So the singular points inside the circle |z| = 1 are 0 and − 12. Note that − 12 is a simple pole. We have
Resz=− 12 z
(^4) − 2 z (^2) + 1 z^2 (2z^2 + 5z + 2)
= z
(^4) − 2 z (^2) + 1 8 z^3 + 15z^2 + 4z
|z=− 12 = · · · =^3 4
For z = 0 which is of order 2, we have
Resz=0^ z
(^4) − 2 z (^2) + 1 z^2 (2z^2 + 5z + 2)
= Resz=
z^4 − 2 z^2 + 2 z^2 +5z+ z^2
= d dz
z^4 − 2 z^2 + 1 2 z^2 + 5z + 2
|z=0= · · · = − 5 4
Therefore∫ (^2) π
0
sin^2 θ 5 + 4 cos θ
dθ = i 4
|z|=
z^4 − 2 z^2 + 1 z^2 (2z^2 + 5z + 2)
dz
= i 4
(2πi(Resz=− 12 z
(^4) − 2 z (^2) + 1 z^2 (2z^2 + 5z + 2)
(^4) − 2 z (^2) + 1 z^2 (2z^2 + 5z + 2)
i 4 (2πi)(
4 +^
4 ) =^
π
(b)
0
x^2 (x^2 + 4)(x^2 + 9) dx. Solution: Let R be a large positive number. Denote by ΓR the line segment on x-axis from z = −R to z = R, by CR the upper half circle {z||z| = R, Imz ≥ 0 } from z = R to z = −R, and by C the simple closed contour C = ΓR
CR.
STEP 1: Compute
C
z^2 (z^2 + 4)(z^2 + 9)
dz. Set (z^2 + 4)(z^2 + 9) = 0 and solve it to obtain z = ± 2 i and z = ± 3 i. So the only singular points of
z^2 (z^2 + 4)(z^2 + 9) inside^ C^ are^ z^ = 2i^ and^ z^ = 3i. Since both of them are simple poles, we have
Resz=2i^ z
2 (z^2 + 4)(z^2 + 9)
= z
2 d dz ((z^2 + 4)(z^2 + 9))^
|z=2i= · · · = i 5
and Resz=3i^ z
2 (z^2 + 4)(z^2 + 9) =^
z^2 d dz ((z^2 + 4)(z^2 + 9))^
|z=3i= · · · = − 103 i.
It follows that (^) ∫
C
z^2 (z^2 + 4)(z^2 + 9) dz^ = 2πi(^
i 5 +^
− 3 i 10 ) =^
π
STEP 2: Show that (^) Rlim→∞
CR
z^2 (z^2 + 4)(z^2 + 9) dz^ = 0. Note that |z| = R when z is on CR. We have
| z
2 (z^2 + 4)(z^2 + 9)
| = |z|
2 |z^2 + 4||z^2 + 9|
≤ |z|
2 (|z|^2 − 4)(|z|^2 − 9)
= R
2 (R^2 − 4)(R^2 − 9)
= M.
Since the length of CR is equal to L = πR,
|
CR
z^2 (z^2 + 4)(z^2 + 9) dz| ≤^ M L^ =^
R^2
(R^2 − 4)(R^2 − 9) πR^ =^
πR^3 (R^2 − 4)(R^2 − 9).
It follows from simple calculus that
πR^3 (R^2 − 4)(R^2 − 9) →^ 0 as^ R^ → ∞. Therefore
lim R→∞
CR
z^2 (z^2 + 4)(z^2 + 9)
dz = 0.
STEP 3: Compute
0
x^2 (x^2 + 4)(x^2 + 9) dx. Since x
2 (x^2 + 4)(x^2 + 9)
is an even function of x, we have ∫ (^) ∞
0
x^2 (x^2 + 4)(x^2 + 9) dx^ =
2 Rlim→∞
ΓR
z^2 (z^2 + 4)(z^2 + 9) dz
=^1 2
( lim R→∞ (
C
z^2 (z^2 + 4)(z^2 + 9)
dz −
CR
z^2 (z^2 + 4)(z^2 + 9)
dz))
=
2 (^
π 5 −^ 0) =^
π
Determine the values of ∆C argf (z) for the function f and positively oriented contour C.
(a) f (z) =
4 z^3 + 3 z ;^ C^ =^ {z^ | |z|^ = 1}. Solution: The function f (z) =
4 z^3 + 3 z has 3 zeros (^
4 )^
(^13) , all of which are inside C = {z | |z| = 1} and of order 1, and it also has one simple pole at 0 inside C. By the Argument Principle, we have
∆C argf (z) = 2π(Z − P ) = 2π(3 − 1) = 4π.
(b) f (z) = tan z; C = {z | |z| = 10}. Solution: Since tan z = sin^ z cos z
, its zeros occur when sin z = 0, and its poles occur when cos z = 0. It follows from simple computations that f (z) = tan z has 7 zeros { 0 , ±π, ± 2 π, ± 3 π} inside C = {z | |z| = 10}, all of which are of order 1, and it has 6 simple poles {±
π 2 ,^ ±^
3 π 2 ±^
5 π 2 }^ inside^ C. By the Argument Principle, we have ∆C argf (z) = 2π(Z − P ) = 2π(7 − 6) = 2π.
- Determine the number of roots, counting multiplicities, of the following equations in the given regions.
(a) z^5 + z^2 + 10z + 3 = 0; D 1 = {z | |z| < 1 }, D 2 = {z | |z| < 2 }, and D 3 = {z | 1 ≤ |z| < 2 }. Solution: 1. D 1 = {z | |z| < 1 }. Let f (z) = 10z, and g(z) = z^5 + z^2 + 3. Both are analytic in D 1. When z is on |z| = 1,
|f (z)| = 10, and |g(z)| = |z^5 + z^2 + 3| ≤ |z|^5 + |z|^2 + 3 = 5.
It follows that |f (z)| > |g(z)| when z is on |z| = 1. By Rouch´e’s theorem, f (z) + g(z) = z^5 + z^2 + 10z + 3 = 0 has the same number of roots, counting multiplicities, as f (z) = 10z = 0 in D 1. Since 10z = 0 has exactly one root 0 of order 1, we conclude that z^5 + z^2 + 10z + 3 = 0 has one root in D 1.
- D 2 = {z | |z| < 2 }. Let f (z) = z^5 , and g(z) = z^2 + 10z + 3. Both are analytic in D 1. When z is on |z| = 2,
|f (z)| = |z^5 | = |z|^5 = 32, and |g(z)| = |z^2 + 10z + 3| ≤ |z|^2 + 10|z| + 3 = 27.
It follows that |f (z)| > |g(z)| when z is on |z| = 2. By Rouch´e’s theorem, f (z) + g(z) = z^5 + z^2 + 10z + 3 = 0 has the same number of roots, counting multiplicities, as f (z) = z^5 = 0 in D 2. Since z^5 = 0 has one root z = 0 of order 5, we conclude that z^5 + z^2 + 10z + 3 = 0 has 5 roots, counting multiplicities, in D 2.
- D 3 = {z | 1 ≤ |z| < 2 }. It follows from the above that z^5 + z^2 + 10z + 3 = 0 has 4 roots, counting multiplicities, in D 3.
(b) 5z^6 = cos z + 1; D = {z | |z| < 1 }. Solution: Let f (z) = 5z^6 , and g(z) = − cos z − 1. Both are analytic in D. When z is on |z| = 1 we have |f (z)| = | 5 z^6 | = 5|z|^6 = 5, and
|g(z)| = | − cos z − 1 | = |
eiz 2 +^
e−iz 2 + 1‖ ≤ |^
eiz 2 |^ +^ |^
e−iz 2 |^ + 1^ ≤^
e 2 +^
e 2 + 1 =^ e^ + 1. Here we have used |eiz^ | = |ei(x+iy)| = |e−y+ix| = e−y^ ≤ e since −y ≤ 1 on the circle |z| = 1; and similarly we have |e−iz^ | ≤ e. It follows that |f (z)| > |g(z)| when z is on |z| = 1. By Rouch´e’s theorem, f (z) + g(z) = 5z^6 − cos z − 1 = 0 has the same number of roots, counting multiplicities, as f (z) = 5z^6 = 0 in D. Since 5z^6 = 0 has one root 0 of order 6, we conclude that 5z^6 − cos z − 1 = 0, or equivalently, z^6 = cos z + 1 also has 6 roots, counting multiplicities, in D.
