The integrals of 1

( )

f z over a,b cycles for horizontal cut First we discuss the value different in sheet-I of theory and on Mathematica .

1in sheet-I,   1 i but 1in Mathematica,  1 i. We will find the mistake in this.

We know that ^{ }



^{ ,}



in Mathematica

Figure 30: Domain and range in Mathematica

Lemma Ⅰ: ^{ }



^{ ,}



in sheet-I for horizontal cut

in Mathematica

2

 2



＋

Now we take a example to test and verify the Lemma Ⅰ.

To evaluate 1

The(1)and(2) difference a ＂－＂ sign.

2. z r₂ mathematica,they difference a ｀ －＇sign.

Now we get the result Lemma Ⅰ,we can use the result to do some example.

Example 1.

2. z r₂

Example 2. Evaluate 1 ( )dz

Figure 32: a-cycle and homotopic path a^ Proof:

(2)theory: za₁^*

(Ⅱ) z^{6 ← 9}

In mathematic the value of

1

2.^z

 

^{3, 4}

(3)mathematic z a₂^

So that

3

(4)by the Lemma (Ⅰ)

b ₁

7 1

In mathematic

3

－ b₂^ － － －

1 1

(Ⅱ) z-1 1

(3)by Lemma (Ⅰ)

*

7 1

(Ⅲ)z 4 6

(Ⅲ)z 4 6

Now we discuss in general situation:

Compute 1

1.a-cycle

Figure 37: a-cycles for 2N-1 points

a _N a_N1 a _j a ₂ a ₁

Figure 38:a-cycles for 2N points

By Cauchy theorem, we can get that

*

Now we consider the path by the Lemma Ⅰ

2 2 1 2 2 1

arg( ) 0 , 2 , 2 1,...,

So we have the conclusion

2 1

b _j

3.Now we consider the equivalent path b_j^ By Cauchy Theorem,we know that

1 1

( ) ( )

j j

b f z  b^ f z

 

The path we must have two part to discuss.

(1) path on the cut is

Now we use LemmaⅠ to computation the question.

Now we use LemmaⅠ to computation the question.

(i)zz_2s1z_2s

(ii) zz_2s1  z_2s　 because f z( )_   f z( )_

3.The integrals of

¹

( )

f z

over a,b cycles for vertical cut

First define that

3 , where z

Each sheet has “＋＂and＂－＂ edge，z is the endpoint of the cut. _k Example:

Because ₂  ₁ 2  where ₁  , ₂ 

Now we discuss the value different between theory and mathematic.

Figure 44：the value of z in theory and mathematic We see that have the problem at 3

2 ,

^^  ^,next we prove that how to solve the problem .

Lemma 2 Take z  for the vertical cut .

(2)when 3 2 ,

 ^  ^

In mathematic ,it will regard as 3

2 , ,

re e re mathematic



Example 1: Evaluate the integrals of 1 ( )

(1) za₁^ 2ii Let z , 2  ri r   1, dz  idr

(2) i 2i Let z , 1  ri r   2, dz  idr

(1) za₂^ 4i3i Let z  , ri r  4  3 , dz  idr

(2) 4i 3i Let z ,  ri r  4  3 , dz  idr

6i

Figure 46: b cycle and homotopic path b^

1.(Ⅰ)theory zb₂^ Let z ,  ri r  5  4 , dz  idr

(Ⅲ) by the Lemma 2 So we consider 4i→5i

, 5 4 ,

(Ⅲ) by the Lemma 2

-b the path along vertical cut from i to i on edge of sheet b the path along vertical cut from i to i on edge of sheet b the path along vertical cut from i to i on sheet



-b^ the path along vertical cut from i to i on sheet



(Ⅱ)mathematic Let z ,  ri r  3  4 , dz  idr

(Ⅲ)by Lemma 2

(Ⅲ)by Lemma 2

Now we give a more easier method to reduce the work process.

3.zC

Next ,we give a example to test the method .

Example Compute 1 _{1 2 1 2}

Figure 48: a-cycle and the equivalent path

1.

a the path vertical cut from i i on edge in sheet a the path vertical cut from i i in edge of sheet



2 1.04083 10 0.119738

( ) (2 )

2.

a the path vertical cut from i i on edge in sheet a the path vertical cut from i i in edge of sheet a the path vertical cut from i i in edge of sheet



a the path vertical cut from i i in edge of sheet a the path vertical cut from i i in edge of sheet a the path vertical cut from i i in edge of sheet



(3) za^*₂₃

(6) za^*₂₆

a the path vertical cut from i i on edge in sheet a the path vertical cut from i i on edge in sheet



(2) za₃₂^*

2 1.73472 10 0.227188

( ) ( )

Next we evaluate the b-cycle .

＋ － b ₁

b ＋ － ＋ － ＋ － ₂ b ₃

Figure 49: the b-cycles

Evaluate _*

b the path horizontal line from i i on sheet b the path horizontal line from i i on sheet

b the path vertical cut from i i in edge of sheet b the path ver

tical cut from i i in edge of sheet b the path vertical cut from i i in edge of sheet b the path vertical cut from i i in edge of sheet b the path v

ertical cut from i i in edge of sheet b^ the path vertical cut from i i in edge of sheet

(2) zb₃₂^ (-1+i -- i ) Let z  r i , r  1 0 dzdr

(5) zb₃₅^ Let z  1 ri , r   1 2 dzidr

By (1) (2) (3) (4) (5) (6) (7) (8),

* 3

2 1 1 2

0 1 2 3

1 2 2 2 2

( ) ( ) ( 1 ) ( 1 ) ( 1 )

math b

i i i i

dz dr dr dr dr

f z f r i f ri f ri f ri

 

  

   

      

    

= 0.0944734 + 0.0000942992i

Evaluate _* b the path horizontal line from i i on sheet b the path horizontal line from i i on sheet

b the path vertical cut from i i in edge of sheet b the path vert

 b the path vertical cut from i i in edge of sheet b the path vertical cut from i i in edge of sheet



(2) zb₂₂^ (i --1+i )

-edge of sheet edge of sheet edge of sheet edge of sheet

    

b vertical cut from i i in edge of sheet b vertical cut from i i in edge of sheet b vertical cut from i i in edge of sheet b vertical cut

Next ,we consider all in sheet-Ⅰ.

zb23^ Let z 1 ri , r 3 2 dzidr

(4) zb₂₄^ Let z 1 ri , r 2 1 dzidr

3. next evaluate _* b the path on a horizontal line from i i on sheet I

b the path on a horizontal line from i i on s

b the path on a vertical cut from i i with edge of sheet II b the path on a vertical cut from i i with edge of sheet II



(3) zb₁₃^ Let z 1 ri , r   2 3 dzidr

(6) zb₁₆^ since f z( )_   f z( )_

-edge of sheet edge of sheet edge of sheet edge of sheet

    

-b vertical cut from i i in edge of sheet b vertical cut from i i in edge of sheet



Next ,we consider all in sheet-Ⅰ.

*

(8) zb₁₈^* Let z 2 ri , r 2 1 dzidr

1 2 1 2

3 4 3 4

5 6 5 6

7 8 7 8

math

math

math

math

z z z z z z z z

z z z z z z z z

z z z z z z z z

z z z z z z z z

      



    



     

     



( ) ( )

math

f z f z

  

Now we discuss the general situation.

Compute over a b for vertical cut where f z z z f z

Figure 52: a-cycle and their equivalent path a^

1.a-cycle: The equivalent path

a =the path on a vertical cut j z_{2 -1j}  z_2j with (+)edge of sheet-I and z_{2 -1j}  z_2jwith (-)edge of sheet-I.

Now we use the Lemma to computation .

(i)z  Z_2j Z_{2 -1j}

math

Now we discuss the two part .

(ii) z  z_2s+1<--- z since _2s f z( )_   f z( )_,

4 The integrals of

¹

( )

f z

over a,b cycle for slant cut

Defnition : The cut with  means the slope of the straight line tan

The same to the horizontal and vertical cut,we define the sheet-Ⅰand sheet-Ⅱ first .

Now we to do the same work , discuss the dfference between the value in theory and in Mathematica.

math

Figure 55:The value in sheet-I

Figure 56: The value in sheet-I and Mathematica Since there have some problem in 7

4 ,

1

1

Figure 58:The value in sheet-I and mathematica

Figure 59:The value in sheet-I and mathematica

Since there have some problem in 4

So that we give a conclusion .

math ＋ － math

 2 2

 2

  Ⅰ 



2 2

 

2

 

Figure 61:Example of f z( ) z

Figure 62: a-cycle and the equivalent path

1. a₁^*a₁₁^* a₁₂^*

1 2 2 2 0.226932 0.0125601

( ) 1 3

2. a₂^*a^*₂₁a₂₂^*

1 2 2 2 0.0584232 0.842766

( ) 3 1

3. a₃^a₃₁^ a₃₂^ ....a₃₆^

(2) ₃₂ ( ^{1 1} ) , 2 2( 3 1) ( ^{1 1} )

(4) ₃₄ ( ^{1 1} ) , 2( 3 1) 0 ( ^{1 1} )

* Figure 63: The path b-cycle

1. b₃^b₃₁^ b₃₂^ ....b₃₆^ b slant line from i on sheet

b slant cut from i on edge of b slant cut from i on edge of sheet b slant line from i on sheet



*

(5)zb₃₆^*

2.b₂^*  a₃₂^* b₂₁^* b₂₂^* b₂₃^* b₂₄^* b₃₁^ b₃₂^* b₃₄^* b₃₅^* b₃₆^*

-b vertical line i i on sheet

b slant cut i i on edge of sheet

b slant cut i i on edge of sheet

b vertical line i i on sheet



(2) ²²

3.b₁^*b₁₁^* b₁₂^* b₁₃^* b₁₄^* b₁₅^* b₃₁^* b₃₅^* b₃₆^*

b horizontal line from i i on sheet

b slant cut from i i on edge of sheet

b slant line from i i on sheet

b horizontal line from i i on sheet

b

(2) _*

arg( ) 5 , , 1, 2

Similarly now we give a more easier method to reduce the work process.

We divided C by many blocks to discuss the slant cuts.

(A) (D)

(4)z ( ) edge of sheet I

So that the conclusion

( ) ( ) ( )

2. For the second example we take 2

(4) ( ) z  edge of sheet I

So that we have the conclusion ( ) ( ) ( )

Now we discuss the general of this special case: for any slant cut which

Figure 69: m=2N-1 for slant cuts

 

(1)

(i) The path on no cuts:^{z  L} 



( , ) :^{x y y  y}1



:

(v)The path on the cut:

When m=2N-1 the conclusion

 

For case Ⅱ: m=2N

Figure 70: m=2N for slant cuts Note that

(1) z  ( )( ) B₁ B₂ the same to the m  2N - 1

(ii) The path on no cuts:^{z  L} 



^{( , ) :x y y2j  y  y2j+1}



(iiii)The path on the cut:



^{j j}



When m=2N the conclusion

 

 

N N

j j j

j j

math

f z if z B z L x y y y y on edge

f z

f z otherwise

 

     



 



-1

2 2 -1 2

1 1

- ( ) ( , ) : ( )

( ) ( )

 

 

Example:

z₈  2i ＋ －

z₆   1 i z₂   4 i z₇  i

＋ － ＋ － ＋ z₃ 2

z₅  0 z₄  － 1

z₁  4 i

Figure 71: The cut of ( )f z

Evaluate 1

( ) and a b , cycle

f z 



( ) ( 4 )( 4 )( 2)( 1)( 0)( 1 )( )( 2 ) let f z  z i z i z z z z i zi z i

＋ － a ₃ a ₁

a^3

＋ － ＋

a^2 a ₂ a1^

_ ＋ －

Figure72 :The cut of f z ( ) in complex plane

1.z a₁ a₁^*a₁₁^* a₁₂^*

a vertical cut from i i on edge of sheet a vertical cut from i i on edge of sheet

       

 

 0.0553983-0.116615i

2. za₂ a₂^*a^*₂₁a₂₂^*

a horizon cut from on edge of sheet a horizon cut from on edge of sheet

     

3.z a₃ a₃^*a^*₃₁a₃₂^*

a slant cut from i on edge of sheet a slant cut from i on edge of sheet

      

b ₁ b ₂

＋ －

b ＋ － ₃ ＋ －

＋ －

Figure 73: the path of b-cycle

1.z b₃^

＋ － b₃^

＋ －

＋ ＋ － －

Figure 74:The path of b₃^

3 3 31 32

-b horizontal line i i on sheet b horizontal line i i on sheet



2.z b₂^ b vertical line i on sheet b horizontal line on sheet b vertical line i on sheet



(2)zb₂₂^

3.z b₁

b horizontal line on sheet

b vertical cut i on edge of sheet b horizontal line i i on sheet



(2)b₁₂^* vertical cut from 4 4to i on ( ) edge of sheet I

在文檔中 N相黎曼空間的理論與應用 (頁 22-131)