Math. 185:
Solutions to Problems due Wed. 9 Apr. 2008
April 9, 2008 4:51 am
Prof. W. Kahan Page 1/3
0.
Exhibit a function ƒ(z)
,
continuous
(but not analytic) throughout the closed unit disk, to
which function ƒ(z) no polynomial can approximate arbitrarily closely thereon.
Answer 0
: Try ƒ(z) := 1 – |z|
2
. It is continuous. But no polynomial P can satisfy
|ƒ(z) – P(z)| < ß < 1/4 throughout the unit disk. This is so because P would have to satisfy
ß > |ƒ(z) – P(z)| = |P(z)| on the disk’s boundary |z| = 1 , and then the Maximum Modulus
Theorem would imply ß > |P(0)| too, forcing |ƒ(0) – P(0)| = |1 – P(0)| > 1 – ß > ß .
Polynomials can easily approximate arbitrarily closely to any function ƒ(z) analytic throughout the
closed
unit
disk. Then the Maclaurin series for ƒ converges in a larger disk of radius bigger than some r > 1 , and
converges in the closed unit disk at least as fast as a geometric series with common ratio 1/r < 1; this means that
the series’ first N terms constitute a polynomial that approximates ƒ as closely as desired if N is chosen big
enough.
The situation is more interesting when ƒ is analytic in the
open
unit disk but merely continuous towards and
along its boundary, or when ƒ is analytic throughout a closed bounded simply connected region with a piecewise
smooth boundary. But those are stories for another day about best polynomial approximation.
1(a).
As z traverses the unit circle
O
once counterclockwise, how many times does z
2
– 1/4
go around the origin? What is
∫
O
2z·dz/(z
2
– 1/4) ?
Answers
: Twice. 4
π
ı
.
1(b).
As
Φ
increases from 0 to 2
π
, the expression (
e
3
ı
Φ
– 1/2)(2
e
–2
ı
Φ
+ 1) traces some
closed curve in the complex plane. How many times does that curve go around the origin?
Answer
: Once because (z
3
– 1/2)(2/z
2
+ 1) has 3 zeros and 2 poles inside the unit circle.
2.
Evaluate
∫
E
z
7
·
e
sin(z)
·dz integrated around the ellipse E traced by z = 5·cos(t) + 2
ı
·sin(t)
as t increases from 0 to 2
π
.
Answer
: 0 , because the integrand is analytic inside and on E .
3.
Evaluate
∫
O
z·
e
z·sin(
π
z)
dz
/
(z – 1/2) integrated once counterclockwise around unit circle
O
.
Answer
:
π
·
ı
·
√
e
= 2
π
·
ı
·{ Residue of z·
e
z·sin(
π
z)
/
(z – 1/2) at z = 1/2 }.
4.
Use residues to evaluate real integrals
∫
–
∞
+
∞
dx
/
(1 + x
2
) and
∫
–
∞
+
∞
dx
/
(1 + x
2
)
2
, showing
what contour you use and which singularities you take into account.
Answers
:
π
and
π
/2 .
The contour is an infinite semicircle above its diameter, the real axis; the singularity is a pole
at z =
ı
; the residue there is –
ı
/2 for the first and –
ı
/4 for the second integral because
2
/
(z
2
+ 1) =
ı
/(z+
ı
) –
ı
/(z–
ı
) and 4
/
(z
2
+ 1)
2
=
ı
/(z+
ı
) –
ı
/(z–
ı
) – 1
/
(z+
ı
)
2
– 1
/
(z–
ı
)
2
.
The next six problems are intended to be solved in sequence.
