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- Find the principal value of
1
2
√
3
2
i
3 i
Solution: Ln
1
2
√
3
2
i
= log e
1
2
√
3
2
i + iArg
1
2
√
3
2
i
π
3
i. So the prin-
cipal value of
1
2
√
3
2
i
3 i
is
e
(Ln
(
1
2
√ 3
2
i
)
)(3i)
= e
π
3
i· 3 i
= e
−π
.
- Evaluate
H
C
7 z− 6
z
2 − 2 z
dz where C : |z| = 4.
Solution:
7 z− 6
z
2 − 2 z
7 z− 6
2
1
z− 2
1
z
Let f (z) =
7 z− 6
2
Then
H
C
f (z)
z− 2
dz = 2πif (2) = 8πi and
H
C
f (z)
z
dz = 2πif (0) = − 6 πi.
So
H
C
7 z− 6
z
2 − 2 z
dz = 8πi − (− 6 πi) = 14 πi.
- Let
H
C
e
z cosz
(z−
π
2
i)
2
dz = a + bi with C : |z| = 2. Find the value of b.
Solution: Let f (z) = e
z cosz. Then
H
C
e
z cosz
(z−
π
2
i)
2
dz =
2 πi
1!
f
′ (
π
2
i). We have
f
′ (z) = e
z cosz − e
z sinz,
so
f
′ (
π
i) = e
π
2
i cos(
π
i) − e
π
2
i sin(
π
i)
= i(
e
−
π
(^2) + e
π
2
e
−
π
(^2) − e
π
2
2 i
= i cosh
π
pi
Thus
a + bi = 2πif
′ (
π
i) = − 2 π cosh
π
π
Hence b = 2π sinh
π
- Solve the problem
u xx
= 0, 0 < x < π, y > 3 ,
u(0, y) = 0, u(π, y) = e
− 2 y
, y > 3 ,
uy (x, 3) = 0, 0 < x < π.
Solution: Let v(x, y) = u(x, y + 3). Then we have
2 v
∂x
2
2 v
∂y
2
= 0, 0 < x < π, y > 0 ,
v(0, y) = 0, v(π, y) = e
−3(y+3) , y > 0 ,
∂v
∂y y=
= 0, 0 < x < π.
The Fourier cosine transform in y is suitable for the problem of v. We define
F
c
{v(x, y)} =
∞
0
v(x, y) cos αy dy = V (x, α).
Then we have
d
2 V
dx
2
− α
2 V = 0.
So we have V (x, α) = c 1 sinh αx + c 2 cosh αx. We also have Fc{ 0 } = 0 and
F
c
{e
−2(y+3) } =
∞
0
e
−2(y+3) cosαy dy
= e
− 6
(−
e
− 2 y
cosαy|
y=∞
y=
α
∞
0
e
− 2 y
sinαy dy)
= e
− 6
(−
e
− 2 y
cosαy|
y=∞
y=
α
e
− 2 y
sinαy|
y=∞
y=
α
2
∞
0
e
− 2 y
cosαy dy)
e
− 6 −
α
2
∞
0
e
−2(y+3) cosαy dy
2 e
− 6
4 + α
2
So c 2
= 0 and c 1
2 e
− 6
(4+α
2 ) sinh απ
. Thus
V (x, α) =
2 e
− 6 sinh αx
(4 + α
2 ) sinh απ
So
v(x, y) = F
− 1
c
{V (x, α)} =
π
∞
0
2 e
− 6 sinh αx
(4 + α
2 ) sinh απ
cos αy dα.
Finally, we have
u(x, y) = v(x, y − 3) =
π
∞
0
2 e
− 6 sinh αx
(4 + α
2 ) sinh απ
cos α(y − 3) dα.
- Assuming that the temperature u is bounded, find the steady-state tempera-
ture u(r, θ) in a plate that coincides with the entire xy-plane in which a circular
hole of radius 2 has been cut out around the origin and the temperature on the
circumference of the hole is f (θ) = 2 + sin 2 θcos
2 θ.
Solution: Assume solutions of the form R(r)Θ(θ). Since
∂
2 u
∂x
2 +^
∂
2 u
∂y
2 = 0, we have
R
′′
R
R
′
rR
Θ
′′
r
2 Θ
= 0. So
r
2
R
′′
R
R
′
R
′′
= λ.
If λ = 0, then Θ = c 1 + c 2 θ. Since Θ has period 2π, we have Θ = c 1. So
R = c 3 + c 4 ln r. Since u is bounded, we have R = c 3. Hence RΘ = A 0.
b) Find a harmonic conjugate function v(x, y) of u(x, y) = x
2 − y
2 − y.
Solution: We have ux = 2x = vy. So v(x, y) =
2 x dy = 2xy + f (x). We also
have uy = − 2 y − 1 = −vx. So vx = 2y + f
′ (x) = 2y + 1. Thus f
′ (x) = 1. It
follows that f (x) =
1 dx = x+c for any constant c. So v(x, y) = 2xy + x + c.
8.a) Evaluate
H
C
z
3 − 1
2 z−i
dz where C : |z| = 1.
Solution: Let f (z) =
z
3 − 1
2
. Then f (
i
2
−i/ 8 − 1
2
. So
I
C
z
3 − 1
2 z − i
dz =
I
C
f (z)
z −
i
2
dz = 2πif (
i
−i/ 8 − 1
· 2 πi =
π
− πi.
b) Evaluate
H
C
dz
z
2 − 1
where C : |z − 1 | =
π
2
Solution: Let f (z) =
1
z+
. Then f (1) =
1
2
. So
H
C
dz
z
2 − 1
H
C
f (z)
z− 1
dz =
2 πif (1) = πi.
