Homework 5
October 14, 2003
3.4 2 Find a solution value problem depicted in Fig. 3.12 and evaluate it at
(0, 0).
Solution: Firstly, we move the original point to 1+i, then we get z = z(1+i).
Then we get two equations as following:
A 0 + B = 10,
A 3 + B =
Homework 9
November 9, 2003
4.6 2 Solution.We use the Cauchy formula. Giving CR a positive orientation
and R 1, we have by generalized Cauchy formula
f (n) (0) =
n!
2i
f ()
d
CR n+1
Therefore
| f (n) (0) |
2
n!
2 Rn (1R)
=
n!
Rn (1R)
1
by using | f (z) |<
Homework 4
October 9, 2003
3.1 15 Using formula(21), compute each of the following residues:
2z+3
(zi)(z 2 +1)
z 3 +4z+9
= (2z+2)(z3)5
(a)
Res(i) for R(z) =
(d)
Res(3i) for R(z)
2z+3
2z+3
(zi)(z 2 +1) = (zi)2 (z+i) . By using formula (21),
2
d 2z+3
(z3)(z
Homework 3
October 1, 2003
2.3 4 Using Denition 4. show that each of the following functions is nowhere
dierentiable.
(a)Rez
(b)Imz
(c) | z |
df
Proof: For (a), we recall the Denition 4, which is dz (z0 ) f (z0 ) =
f (z0 + z)f (z0 )
1
i
lim z0
. Fix z0 ,
Homework 2
September 24, 2003
1
1.6 6 Describe the boundary of each of the given sets.
(a) | z 1 + i | 3
(c)0 <| z 2 |< 3
(e) | z | 2
(b) | Argz |<
4
(d) 1 < Imz 1
(f )(Rez)2 > 1
For (a), in fact, it is a disc. with original point 1 i and ridus 3, that m
Homework 12
December 1, 2003
6.1 1 Solution. For (a), the singularity of this function is 2. And because the
3z6
function equals e6 ez2 , therefore a1 = e6 . For (d), only one singularity, -1,
6
2
and the function equals (1 z+1 )3 = 1 z+1 + . Therefore, R
Homework 13
December 7, 2003
6.4 1 Solution. We do it by using the same method as Example 1 in 6.4. We
write it as
I = I1 + I2 = p.v.
ei2x
2(x2 +1)
+ p.v.
ei2x
2(x2 +1)
The singularities are i and i. For I1 , we deal with
ei2z
2(z 2 +1)
f1 (z) =
We enco
Homework 11
November 24, 2003
5.3 2 Solution. According to Theorem 2 in page 5.1, we let cj = aj (z z0 )j .
Then by Thm 2, we get that lim | cj+1 /cj |= lim | aj+1 /aj (zz0 ) |= L | zz0 |.
If the limit is less than 1, then converge. That means if | z z0 |
0.1.
Exercise #2. f (x, y) = e
x
sin y
x(t) = t, y(t) = t
Let w(t) = f (x(t), y(t). Then
3
dw
f dx f dy
=
+
dt
x dt
y dt
= (ex sin y) 1 + (ex cos y) 3t2
= et sin t3 + 3t2 et cos t3
and
dw
(1) = e sin 1 + 3e cos 1
dt
0.2.
Exercise #8.
dw
w dx w dy
=
+
dt
x
Homework 8
November 2, 2003
4.4 9 Solution. For (a) obviously, it is simply connected. For (b), No. Because given the circle | z |= 1.5 the interior of it is not wholly in D. For (c),
Yes. For (d), Yes. And for (e), No, because if yes, the we can nd a poi
Homework 10
November 17, 2003
5.1 1 Solution. (c),
j 2 j
j=0 (1) ( 3 )
1
1( 1 )
3
by using Lemma 1. We can get the sum is
For (d),
1
j=0 [ j+2
. We get that it is equal to
1
2
1
1
+
1
3
+
1
2
= 3.
5
1
j+1 ]
+ = 1
1
1
5.1 2 Solution. (a), the cj+1 /cj = (j
Homework 7
October 26, 2003
4.1 4 Show that the range of the function z(t) = t3 + it6 , 1 t 1, is a
smooth curve even though the given parametrization is not admissible.
Proof: Firstly, it is easy to see the curve is just as the function f (x) = x2 , 1
x