... z
z
3
z=−ı
+
2
2!
d
2
dz
2
(z
3
+ z + ı) sin z
z + ı
z=0
= 2 (−ı sinh(1)) + ıπ
2
3z
2
+ 1
z + ı
−
z
3
+ z + ı
(z + ı)
2
cos z
+
6z
z + ı
−
2( 3z
2
+ 1)
(z + ı)
2
+
2( z
3
+ z + ı)
(z ... formula.
C
z
z
2
+ 1
dz =
C
1 /2
z −ı
dz +
C
1 /2
z + ı
dz
=
1
2
2 +
1
2
2
= 2
3.
C
z
2
+ 1
z
dz =
C
z +
1
z
dz
=
C
z dz +
C
1
z
dz
= 0 + 2
= 2...
... cos x
=
0
2
= 0
lim
x→0
csc x −
1
x
= 0
109
Solution 3.15
a.
f
(x) = ( 12 −2x)
2
+ 2x( 12 − 2x)( 2)
= 4( x −6)
2
+ 8x(x − 6)
= 12( x 2) (x − 6)
There are critical points at x = 2 and x = 6.
f
(x) ... 6.
f
(x) = 12( x 2) + 12( x − 6) = 24 (x 4)
Since f
(2) = 48 < 0, x = 2 is a local maximum. Since f
(6) = 48 > 0, x = 6 is a local minimum.
b.
f
(x...
... ations are
u
x
= v
y
and u
y
= −v
x
−(x − 1)
2
+ y
2
((x − 1)
2
+ y
2
)
2
=
−(x − 1)
2
+ y
2
((x − 1)
2
+ y
2
)
2
and
2( x − 1)y
((x − 1)
2
+ y
2
)
2
=
2( x − 1)y
((x − 1)
2
+ y
2
)
2
The Cauchy-Riemann ... 8 .4. 3 1/ sin (z
2
) has a second order pole at z = 0 and first order poles at z = (nπ)
1 /2
, n ∈ Z
±
.
lim
z→0
z
2
sin (z
2
)
= lim
z→0
2z...
... =
1
2
Log
x
2
+ y
2
+ ı Arctan(x, y).
4 52
2. We calculate the first partial derivatives of u and v.
u
x
= 2
e
x
2
−y
2
(x cos(2xy) − y sin(2xy))
u
y
= 2
e
x
2
−y
2
(y cos(2xy) + x sin(2xy))
v
x
= ... x
direction.
f
(z) = u
x
+ ıv
x
f
(z) = 2
e
x
2
−y
2
(x cos(2xy) − y sin(2xy)) + 2
e
x
2
−y
2
(y cos(2xy) + x sin(2xy))
f
(z) = 2
e
x
2
−y
2
((x +...
... n)
n
11.
∞
n =2
(−1)
n
ln
1
n
12.
∞
n =2
(n!)
2
(2n)!
13.
∞
n =2
3
n
+ 4
n
+ 5
5
n
− 4
n
− 3
5 62
Im(z)
Re(z)
R
R
2
1
Im(z)
Re(z)
R
R
2
1
C
r
1
r
2
z
C
C
C
1
2
z
Figure 12. 5: Contours for a Laurent ... closed form. (See Exercise 12. 9.)
N−1
n=1
sin(nx) =
0 for x = 2 k
cos(x /2) −cos((N−1 /2) x)
2 sin(x /2)
for x = 2 k
The partial sums have infinite...
... π)
Hint 12. 23
CONTINUE
Hint 12. 24
CONTINUE
Hint 12. 25
Hint 12. 26
Hint 12. 27
Hint 12. 28
Hint 12. 29
Hint 12. 30
CONTINUE
581
Solution 12. 22
cos z = −cos(z − π)
= −
∞
n=0
(−1)
n
(z −π)
2n
(2n)!
=
∞
n=0
(−1)
n+1
(z ... polynomial.
2 6 12 20
4 6 8
2 2
We s ee that the polynomial is second order. p(n) = an
2
+ bn + c. We solve for the coefficients.
a + b + c = 2
4a + 2b + c...
... the
residue by expanding the function in a Laurent series.
(1 − cos z)
2
z
7
= z
−7
1 −
1 −
z
2
2
+
z
4
24
+ O
z
6
2
= z
−7
z
2
2
−
z
4
24
+ O
z
6
2
= z
−7
z
4
4
−
z
6
24
+ O
z
8
=
1
4z
3
−
1
24 z
+ ... −
n−1
k=0
lim
z→
e
ıπ(1+2k)/n
log z + (z −
e
ıπ(1+2k)/n
)/z
nz
n−1
= −
n−1
k=0
ıπ(1 + 2k)/n
n
e
ıπ(1+2k)(n−1)/n
= −
ıπ
n
2
e...
... are,
Res
1
z
4
+ 1
,
e
ıπ /4
= lim
z→
e
ıπ /4
z −
e
ıπ /4
z
4
+ 1
= lim
z→
e
ıπ /4
1
4z
3
=
1
4
e
−ı3π /4
=
−1 − ı
4
√
2
,
Res
1
z
4
+ 1
,
e
ı3π /4
=
1
4(
e
ı3π /4
)
3
=
1
4
e
−ıπ /4
=
1 − ı
4
√
2
,
We evaluate ... Res
z
2
(z
2
+ 1)
2
, z = ı
Res
z
2
(z
2
+ 1)
2
, z = ı
= lim
z→ı
d
dz
(z − ı)
2
z
2
(z
2
+ 1)
2
= lim
z→ı
d
dz...