Just a concise solution
Author: One knows nothing
Date: June 10, 2022
Stay hungry,Stay foolish
Example 0.1 Find the analytic function f (z) as following with the real part u(x, y)
1. u(x, y) = x2 − y 2 + xy, f (0) = 0
2. u(x, y) = x3 − 3x2 y, f (0) = 0
Solution 1. Just refer to the Cauchy − Riemann equation:
∂u
∂v ∂u
∂v
=
=−
∂x
∂y ∂y
∂x
we can know that
ˆ
∂v
∂v
dx +
dy
v(x, y) =
∂x
∂y
ˆ
∂u
∂u
dx
= − dy +
∂y
∂x
ˆ
= (2y − x)dx + (2x + y)dy
ˆ
1
= d( (y 2 − x2 ) + 2xy)
2
1 2
= (y − x2 ) + 2xy [f or f (0) = 0]
2
(1)
(2)
then,f (z) = u(x, y) + iv(x, y) = x2 − y 2 + xy + i( 12 (y 2 − x2 ) + 2xy) = (1 − 2i )z 2 .
2. Similarly ,
ˆ
ˆ
v(x, y) =
6xydx + (3x2 − 3y 2 )dy =
therefore,f (z) = (x + iy)3 = z 3
Example 0.2 Calculate integrals as following:
1.
˛
|z|=2
Solution
I=
1
[
4
˛
1
dz −
z−1
˛
d(3x2 y − y 3 ) = 3x2 y − y 3
(3)
1
dz
z4 − 1
1
1
dz] − [
z+1
4i
˛
1
−
z−i
˛
1
]
z+i
Due to the important integral formula when a is in the integral region,
{
˛
2πi n = 1
1
dz =
n
0 else
l (z − a)
(4)
(5)
we have that I = 0.
˛
2.
|z|=1
z
dz
(2z + z)(z − 2)
Solution Due to the Cauchy integral formula or Residue theorem (suggested):
˛
z
z
1
1 =
dz = 2πi
I=
|
πi
2(z − 2) z=− 2
5
2(z + 12 )(z − 2)
3.
˛
|z|=2
Solution
I = 2πi[
sin( π4 z)
dz
z2 − 1
√
sin π4
sin(− π4 )
+
] = 2πi
2
−2
˛
4.
|z|=1
Solution
I=
(6)
(7)
ez
dz
z3
2πi z ′′
[e ] |z=0 = πi
2!
(8)
5.
˛
2z 2 − z + 1
dz
(z − 1)3
|z|=3
Solution
I=
2πi 2
[2z − z + 1]′′ |z=1 = 4πi
2!
(9)
Example 0.3 Calculate integrals as following:
´ 1+i
1. 1 zez dz
´i
2. 0 3ez + 2zdz
Solution Obviously,both of then are analytic function,we can calculate them like Riemann integrals:
I1 = ez (z − 1)|1+i
= ie1+i
1
Example 0.4 Calculate the integral
Solution Let x = t,then y = t2 ,
(10)
I2 = (3ez + z 2 )|i0 = 3ei − 4
´
l
Rezdz by the line of y = x2
ˆ
I=
ˆ
1
f (z)dz =
t(1 + i2t)dt =
0
1 2
+ i
2 3
(11)
Example 0.5 Calculate the convergent radius of series as following:
1.
∞
∑
1
(z − i)k
k
k=1
Solution
R = lim
k→∞
2.
ak
k+1
=1
= lim
k→∞
ak+1
k
(12)
∞
∑
z
( )k
k
k=1
Solution
√
1
k
R = lim √
kk = ∞
=
lim
k→∞ k ak
k→∞
3.
∞
∑
k=1
Solution
z
k!( )k
k
k
k!( k1 )
k + 1k
=
lim
(
)=e
k→∞ (k + 1)!( 1 )k+1
k→∞
k
k+1
R = lim
4.
(13)
∞
∑
(14)
k k (z − 3)k
k=1
1
=0
R = lim √
k→∞ k k k
Example 0.6 Expand the functions as following into Laurent series:
1.
1
f (z) =
| z |> 3
(z − 2)(z − 3)
2
(15)
Solution
1
(z − 2)(z − 3)
1
1
=
−
z−3 z−2
1
1
1
1
= (
)− (
)
z z − z3
z z − z2
∞
1 ∑ 3
2
= [ ( )k − ( )k ]
z
z
z
f (z) =
(16)
k=0
=
=
1
z
k=0
∑
1
1
[( )k − ( )k ]z k
3
2
−∞
k=−1
∑
−∞
2.
f (z) =
1.
1
1
[( )k+1 − ( )k+1 ]z k
3
2
1
1 <| z |> 2 and | z |> 2
z − 3z + 2
1
1 1
−
(1 <| z |> 2)
− 1 z 1 − z1
∞
∞
1∑ z k 1∑ 1 k
( ) −
( )
=−
2
2
z
z
f (z) =
1
2
z
2
k=0
=−
(17)
k=0
k=−1
∞
∑
1 ∑ z k+1
zk
( )
−
2
2
−∞
k=0
2.
1
1
1 1
(
)−
2
z 1− z
z 1 − z1
∞
1∑ 2 k
1
=
[( ) − ( )k ]
z
z
z
f (z) =
| z |> 2
k=0
=
k=−1
∑
−∞
Example 0.7 Find the singular points as following:
1.
Solution f (z) =
ez
(z+i)(z−i) ,then
1
[( )k+1 − 1]z k
2
ez
z2 + 1
z = ±i is first order pole.
2.
1
ze z
Solution f (z) = z[1 +
1
z
+
1
2z 2
3.
+
1
6z 3
+ · · · ],z = 0 is Essential Singularity.
1 − cos z
z2
Solution f (z) =
1−cos z
|z→0
z2
= 21 ,z = 0 is Removable Singularity.
4.
sin z
z5
3
Solution f (z) = z−z z/3!+···
,which means the highest negative power of f (z) is −4.f (z) is 4th order pole.
5
Example 0.8 Find the residues of f (z) as following:
3
(18)
1.
ez
z+1
z
z
e
e
Solution z = −1 is first order pole and Res( z+1
, −1)[(z + 1) z+1
]|z=−1 = 1e
1
z = ∞ is essential singularity and Res(f (z), ∞) = − e (The residue of ∞ is the negative of the sum of the other
residues. )
2.
z
(z − 1)(z − 2)2
z
Solution z = 1 is first order pole and Res(f (z), 1) = [ (z−2)
2 ]|z=1 = 1
z ′
z = 2 is second order pole and Res(f (z), 1) = [ z−1 ] |z=2 = −1
z = ∞ is removable singularity and Res(f (z), ∞) = 0
3.
ez
z 2 + a2
ez
(z+ia)(z−ia) ,then z
e−ia
eia
e−ia
−2ia ,Res(f (z), ∞) = −[ 2ia + −2ia ]
Solution f (z) =
= ±ia is the first order pole and Res(f (z), ia) =
=
,Res(f (z), e−ia ) =
− sin a
a
4.
eiz
z 2 + a2
−a
a
e
Solution Similarly,Res(f (z), ia) = e2ia , Res(f (z), −ia) = 2ia
, Res(f (z), ∞) =
Example 0.9 Calculate the integration as following:
1.
˛
cos z
3
|z|=1 z
Solution Due to cos z = z −
z2
2
sinh a
ias
+ · · · ,the a−1 of f (z) =− 12 ,thus I = 2πiRes(f (z), 0) = −πi
˛
2.
1
e z2
|z|=2
Solution Similarly,the a−1 of f (z) = 0,I = 0
˛
3.
|z|=1
Solution f (z) =
4.
eia
2ia
2z+4
z(2z−1)
=
2z+1
,then
2z(z− 21 )
2z + 1
2z 2 − z
I = 2πi[Res(f (z), 0) + Res(f (z), 12 )] = 2πi
˛
|z|=3
ez
(z − 1)(z + 3)2
[z = −3 is out of the integration region.]
Solution I = 2πiRes(f (z), 1) =
∑∞
2
Example 0.10 ¹ Expand f (x) = x + x into Fourier series,then prove that k=1 k12 =
πie
8
¹The process of integration is complex ,please be patient and careful.
4
π2
6
Solution
ˆ
1 π
An =
(x + x2 ) cos nxdx
π −π
ˆ
1 π 2
=
x cos nxdx
π −π
ˆ π
2
=−
x sin nxdx [f orintegrationbyparts]
πn −π
4
= 2 (−1)n
n
ˆ π
1
π2
A0 =
(x + x2 )dx =
2π −π
3
ˆ
1 π
Bn ==
(x + x2 ) sin nxdx
π −π
ˆ
1 π 2
=
x sin nxdx
π −π
2
= (−1)n+1
n
Therefore,
(19)
∞
f (x) =
π2 ∑
4
2
+
[(−1)n 2 cos nx + (−1)n+1 sin nx]
3
n
n
n=1
∑∞
2
Let x = π,,f (x) = π3 + n=1 n42 = 21 (f (π) + f (−π)) = π 2 (for Dirichlet’s theorem)
∑∞
2
Thus, k=1 k12 = π6
Example 0.11 Find the Fourier Transform of f (t) as following.


0 t<0


f (t) = kt 0 < t < T



0 t>T
Solution
F {f (t)} = F (ω) =
1
2π
=
1
2π
ˆ
+∞
(21)
f (t)e−iωt dt
−∞
ˆ T
kte−iωt dt
0
ˆ T
1 −1
k
−iωT
=
kT e
+
e−iωt dt
2π iω
2πiω 0
kT i −iωT
k
=
e
+
[e−iωT − 1]
2πω
2πω 2
k
e−iωT − 1
=
(iT e−iωT +
)
2πω
ω
Example 0.12 Solve the PDE as following:
(20)
 2
2
∂ u
2∂ u


=
a

 ∂t2
∂x2
u|x=0 = 0, u′ |x=l = 0




u|t=0 = φ(x), u′ |t=0 = ψ(x)
Solution Let u(x, t) = X(x)T (t),then we have a2 X ′′ (x)T (t) = X(x)T ′′ (t),i.e.
X ′′ (x)
T ′′ (t)
= 2
= −λ
X(x)
a T (t)
let λ > 0,or the solution of the u(x, t) would be trivial.Then we have
X ′′ (x) + λX(x) = 0
5
(22)
(23)
(24)
(25)
consider k as
√
λ,we get the solution :
X(x) = A cos kx + B sin kx
due to X(0) = 0,we have A = 0.
On the other hand,X ′ (L) = 0,which means Bk cos kx = 0,i.e.k = 2n+1
2L π, n = 0, 1, 2, · · ·
2n + 1
Xn (x) = Bn sin(
πx) n = 0, 1, 2 · · ·
2L
And
a(2n + 1) 2
T ′′ (t) + (
π) T (t) = 0
2L
Whose solution is
a(2n + 1)
a(2n + 1)
Tn (t) = cn cos
πt + dn sin
πt n = 0, 1, 2 · · ·
2L
2L
Thus,we get the solution:
a(2n + 1)
a(2n + 1)
2n + 1
un (x, t) = [Cn cos
πt + Dn sin
πt] sin
πx n = 0, 1, 2 · · ·
2L
2L
2L
Then,u(t = 0) = Cn sin 2n+1
2L πx = φ(x),Cn is solved by fourier expansion:
ˆ
2 L
2n + 1
Cn =
φ(x) sin
πxdx n = 0, 1, 2 · · ·
L 0
2L
Similarly,u′ (t = 0) =
sin 2n+1
2L πx = ψ(x),
ˆ L
2n + 1
4
ψ(x) sin
πxdx n = 0, 1, 2 · · ·
Dn =
(2n + 1)πa 0
2L
Finally,due to the superposition principle
∞
∑
a(2n + 1)
a(2n + 1)
2n + 1
u(x, t) =
[Cn cos
πt + Dn sin
πt] sin
πx
2L
2L
2L
n=0
(26)
(27)
(28)
(29)
(30)
(31)
2n+1
2L aπDn
Cn and Dn is determined by 31 and 32
6
(32)
(33)