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Math 241 Exam 2 Solutions - Rimmer
(^ )^
(^ )
1. Solve using an appropriate Fourier Transform
1,^0
,^ 0,
0 with conditions
xx^ t^
x
x
ku^
u^ x^
t^
u^ t
u^ x^
<^ < x
=^
>^
>^
=^
=^ ^
(^ )
0, and
0 given
use Fourier Cosine Tran
sform
x
x^
u^ t
>^
=^
{^
}^
{^ }
C^
xx^
C^ t
F^ ku
F^ u = (^ ){ }
( )^
(^ ){ }
2
0
,^
0,^
C^
x^
d C
k^
F^ u^
x t^
u^ t
F^ u^
x t
dt
^ α
^
^
^
−^
−^
=^ ^
^
^
(^ ){ }
Let^
F^ F^ u x t =^ C
dF 2 k F
dt
α− =
2
dF^
k^ dt
F
α= −^2
dF^
k^ dt
F
α =^ −∫ ∫ ( )
2
ln^ F^
k^ t^
C α
= −^
(^ )^
2 ln^ F^
k^ t^ C
e^
α− + e =^2^ k^
t
F^ Ae
α−
(^ ){ }
2
,^
k^ t
F^ u^ x t^ C
Ae
α−
, 0 F u x (^) ( ){ } C
A =
, 0 F u x (^) ( ){ } C
A =
(^ ){ }
(^ ) 2 0 , 0^
1 cos F^ u^ xC
x dx α =^ ⋅∫ (^ ){ }
(^ )^
(^ ) 2 0 sin^
sin 2
, 0 C
x
F^ u x
α
α α
α
=^
= sin 2^ (^ ) A
α α = ( ) {^
}^
(^ )^
2
sin 2
,^
k^ t
F^ u^ x t^ C
α^ α^ − e α
(^ )^
(^ ){ } 1 ,^
, C^
C u x t^
F^
F^ u^
x t −^ ^
=^
^
(^
)^
(^
)^
(^
) 2
sin 2 2 0 ,^
k^ t cos
u x t
e^
x d
α^
α^
α
π^
α ∞
=^
∫
Math 241 Exam 2 Solutions - Rimmer
- Evaluate^134
3 :^
2 C
dz z^
z C^ z^
(^
)
3
2 4
4
z^
z^ z^
z +^ =
+^
(^
)^ (^
) 2
2 z^ z^
i^ z^
i
=^
−^
(^
)(^
)
(^1 )
1 4
2
2
z^
z^ z
z^
i^ z^
i
= +^
−^
+^
2
2
A^
B^
C z^ z
i^
z^ i =^ +
:^0
1 A^ z^ = A^^ =^ z
(^
)(^
(^ ) 1 )(^ )
1 2 2
4
0 2
0 2
i^ i
i^
=^ i
= −
−^
(^
) :^2
1 2
2 B^ z^
i B^
i^ z^
i = = −
(^
)
1
1 2 4
8
2 2
i^ i i^ i
− =^
=⋅
(^
) (^
)
:^
2
1 2 2
2
2
C^ z^
i B^
i^ i
i^
z^ i = − = −^ −
−^
+^
1 (^ )
1 2
4
8 i^
− i
=^
= −^ ⋅ −
1
1 1
8
8 4 (^1 )
3 :
4
2
2
2
C^
C dz^
dz^ C
z^
i
z^
z^
z^ z^
i^ z^
i −^
−
^
=^
+^
+^
−^ =
^
+^
−^
^
∫^
∫
1
1
1
8
8
4
singularity
singularity inside
outside 2
2
C^
C C^
C dz^
dz
z^ z^
i^
z^ i −^
−
^
=^
+^
^
−
^
∫^
∫
1
2 1
1 4
8 1
1
0 2
C^
C dz^
dz
z^
z^ i
=^
−^
− ∫^
∫
1
2 1
1 :^ and
:^24
4
C^ z^
C^ z^
i =^
−^ =
(^
) 12 8
i π
(^ )^^ i^ π^ =^4
(^
)
2
2 i^
i π
π
=^
−
Math 241 Exam 2 Solutions - Rimmer
)^
(^ )
)^
(^ )
- TRUE/FALSE. Decide on the truth val
ue of the following statements.
An explanation is required in each case.
I^ The set of points given by the equati
on^1
can be described as Re
II^ If^
z^
z^
z
f^ z
−^ ≥^
≤^
(^ )
)
on the curve
given by
3, then
III^
for all^
C
z^ z
C^
z^
f^ z dz
e^ e^
z
π
=^
∫
I^ )
Let
z^
z^
z^ x
iy
−^ ≥
=^ +
2
2
2
2
x^
x^
y^
x^
y
−^
+^ +^
≥^
(^
(^2) )
2
2
2
1 x
y^
x^
y
−^
+^
≥^
(^
(^2) )
2
2
2
1 x
y^
x^
y
−^
+^
≥^
1 0 x
−^ +
≥ (^ )
Re^
x^
z
≤^ ⇒
1 x ≤
T
)^
(^ )^
(^ )
II^ If^
2 on the curve
given by
3, then
C
f^ z^
C^
z^
f^ z dz
π
≤^
=^
∫
(^ )
2,^
C
M^
L^
r^
f^ z dz
π^
π
π
=^
=^
=^ ⇒
∫^
T
III^ Let^ )
z^ ,
x^ iy
z^ x
iy e
e
−
=^ +
= ( )^
(^ )
(^
)
cos^
sin
x^ iy^
x e^
e^
y^ i
y
−^ =^
−^ +
− (^ )^
(^ )
(^
)
cos^
sin x e
y^ i
y
=^
− (^ )^
(^ )
cos^
sin
x^
x
e^
y^ ie
y
=^
−^
z e =
T
Math 241 Exam 2 Solutions - Rimmer
(^ )^
(^ ) (^
)
- Consider a semicircular plate of rad^ (^ )
ius 1 whose boundary on the
is insulat
ed
, 0^
,^0
and has its circular edge maintained at the temperature 1,^ 6 cos 2
5cos 3. The steady-state temperatu u^ r^
u^ r θ u
πθ θ
θ
θ =^
=^
−^
(^ )^
1 12
re^ ,^
satisfies
Find^
,^. 2 6
rr^ rr^
r
u r^
u^ u
u
u
θθ
θ
π
+^ +
^ ^ ^ ^ (
)^
(^ )^ (
Assume :
, u r^
R r θ
θ
=^
2
r R^
rR^
R λ λ
′′^ ′ +^
−^ =
^
′′Θ +^ Θ =
( ) 1
0 2 c^
c θ^
′′⇒ Θ =θ Θ^
=^
λ^ = 0^ (^ ) 0 1
(^0) c ′Θ =
= (^ )^
c^2 θΘ = ( )
2
2
3
4 say^
0
cosh^
sinh c^
c λ^
α
α
θ
θ
′′ θ
<^
= −^
⇒^ Θ −
Θ =
Θ^
=^
λ^^0 Hyperbolic Sine and Hyperbolic Cosineare not periodic. For this
the only pe
riodic
solution is the trivial solution.
λ
(^ )^
(^ )
(^ ) 2
2
5
6
say^
0
cos^
sin
c^
c λ^ α
α
θ
αθ
αθ ′′
>^
=^
⇒^ Θ +
Θ =
Θ^
=^
λ^^0^ (^ )
6
6
0
0
0 c^
c α ′Θ =
=
⇒^
=
(^ )^
(^ )^
(^ )
5
6 sin^
cos
c^
c
θ^
α^
αθ^
α^
αθ
′Θ = −
(^ )^
cos(^ ) 5 c θ
αθ Θ^
= ( )^
sin^ (^ ) 5
0 c π^
α^
απ
′Θ = −
= ,^ 1, 2,3 n n α⇒ =^
=^
… 2 ,^
1, 2, n^ n λ ⇒^
=^
=^
…
(^ )^
cos(^ ) 5 c^
n θ
θ
Θ^
=
Math 241 Exam 2 Solutions - Rimmer
)^1 )^24 )^12
2
(^12)
- Evaluate the following contour integr^2
als, express the answer in
form.
1 where
is the unit circle where^
is the straight line segment joining 1 to 1 C z C i z
a^ bi
a^ z^
dz^
C
b^ e dz
C^
i
c^ ze^ π dz
∫ ∫ ∫^ )^ (^
) ( ) ( ) (^21)
(^12) 2
2 2 2
2
(^1 ) where^
is the unit circle cos^
sin 1 cos
sin 1 cos
sin 1 cos
2 cos
1 sin 1 2 1
cos a^ z^ C
dz^
C
z^ t
i^ t z^
t^ i
t z^
t^
t
z^
t^
t^
t
z^
t − = + − = −^
−^ =^
−^ +
−^ =^
−^
+^ +
−^ =^
∫
(^
) sin^
cos dz^
t^ i^
t dt =^ −^
(^
)(^
)
cos^
sin^
cos
z^ dz
t^
t^ i^
t dt
−^
=^ −^
−^ +
(^
2
2
2 sin^
sin cos
cos
cos
z^ dz
t^ t^
t^ i^
t^ i^
t dt
−^
=^ −^
+^
+^
(^
1
2 2
2
0 1
2 sin
sin cos
cos
cos
z^ C dz^
t^ t^
t^ i^
t^ i^
t dt
π −^
=^ −
+^
+^
∫^
∫^222 cos i^^0 π tdt = −^ ∫
(^
) 212 1 0
cos 2 2 i^
t dt π = − −∫ (^
) 2 1 cos 2 i^0
t dt π = − −∫^2 i^ π= −
) 2
2 The Fundamental Theorem of Contour Integralswhere^11
is the straight line segment joining 1 to 1 b z^ e dz^ C iz
C^
i
π e
∫ =
1
(^1) i e^ π+ e =^
−^
(^ i^1 ) π e e = −^
2 e = −
i 42 ) 12
z
c^ ze
π^ dz ∫^2212214
z^ z^ z
D^ I^ z^ e^ e^ e
(^412) 2
2 1
1 2
4
i z^
z
ze^
π e
^
=^
^
^
[^
]
2 1 12
1 8
4
4 4 i^
i
ie^
e^
e^ e
π
π
=^
−^
−^ −
^
12 4 2
1 π^^ i π e i =^
− 1 4 2
1 π i i = − ^1 π i 8 4 =^ −
−
Math 241 Exam 2 Solutions - Rimmer
(^
- Answer the following questions.a) Find all solutions to the equation
3
3 3
b) Find the principle value of
1
3
in^
form. z^ i
i i^
a^ bi =^
− +^
5
a) Find all solutions to the equation
3
3 3. z^
i =^
−
(^
3 3
3 z^
i =^
− 6, 6 r
π θ= = − 1 The^
roots of
are^2
cos^
sin^
0,1, 2,^
th n nk
z k^
k
w^ r^
i^
k^
n
n^
n
θ^ π
θ^ π
^
+^
+^
^
^
^
=^
+^
=^
^
^
^
^
^
^
^
^
^
(^5) n =^15 6
0 0 :^5
6 cos
sin 30
k^
w^
i
π
π
π
−^ π
^ −
−^
^ ^
^
=^
+ ^
^
^ ^
^
=^
=^
15
6
1
6 cos
sin 3
2 11 1:^5
w^
i
k
π
π
π
−^ π^ π
^
^ ^
^
=^
^ ^
^
^
^ ^
^
=^
=^
⇒^
15
6
2
6 cos
sin 3
2 :^5
w^
i
k
π
π
π
−^ π^ π
^
^ ^
^
=^
^ ^
^
^
^ ^
^
=^
=^
⇒^
15
6
3 6 35
6 cos
sin 6
3 :^5
w^
i
k
π
π
π
−^ π^ π
^
^ ^
^
=^
+ ^
^
^ ^
=^
=^
⇒^
15
6
4
6 cos
sin 3
4 :^5
w^
i
k
π
π
π
π^
π −^
^
^ ^
^
=^
^ ^
^
^
^ ^
^
=^
=^
⇒^
(^
b) Find the principle value of
1
3
in^
form. i i a
bi +^
(^
)^
Ln 1^3 (^ ) 2
2 1 3
i^
i^
i i^
e^
+^
=
(^
Ln 1^
3 ln 2
3 i^
π i +^
=^
(^
Ln 1^
ln 2
i^
i i
π i
^
+^
=^
(^
Ln 1^
ln^2
i^
i^
π i
+^
= −^ +
(^ )^
ln^2 Ln 1^32
6 i^
i i e^
π − + e +^ =
π −ln^2 i^6 e e
(^
)^
(^
)^
(^
6
6
2
cos l
n^2
sin ln
i
e^
i
i^
e
π
π
−^
+^
