COMPLEX ANALYSIS: Mathematica 3.0 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 1998
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
3rd Edition,  1997,  ISBN: 0-7637-0270-6
Jones & Bartlett Publishers, Inc.
40 Tall Pine Drive, Sudbury, MA 01776
Tele.  (800) 832-0034, FAX:  (508)  443-8000
E-mail:  mkt@jbpub.com,     http://www.jbpub.com/
This free software is compliments of the authors.
mathews@fullerton.edu,  howell@westmont.edu

CHAPTER 6  COMPLEX INTEGRATION
Section 6.1  Complex Integrals
Section 6.2  Contours and Contour Integrals
Section 6.3  The Cauchy-Goursat Theorem
Section 6.4  The Fundamental Theorems of Integration
Section 6.5  Integral Representations for Analytic Functions
Section 6.6  The Theorems of Morera and Liouville and Some Applications
GoTo Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.
GoTo Contents

CHAPTER 6  COMPLEX INTEGRATION

Clear[c, d, f, F, i, int, int1, int2, soset, solset, t, u, U1, U2, v, 
      V1, V2, val, w, W, W1, W2, Wint, x, x0, x1, y, y0, y1, z, z1, z2, Z, 
      Z1, Z2, Z3];

Section 6.1  Complex Integrals

In Chapter 3 we saw how the derivative of a complex function is defined.  
We now turn our attention to the problem of integrating complex functions.  
For the integrand  f(t)  =  u(t)  +  i v(t), we define the complex integral:
     ∫ f(t) dt   =  ∫ u(t) dt  +  i∫ v(t) dt .

Example 6.1,  Page 157.  Show that   (∫_( 0)^( 1))(((t  -  i))^3) 
              dt  =  (( - 5)/4).

\(Clear[f, int, t, u, v, val]; 

    f[t_]\  = \ ComplexExpand[\((t\  - \ I)\)\^3]; 

    u[t_]\  = \ ComplexExpand[Re[f[t]]]; 

    v[t_]\  = \ ComplexExpand[Im[f[t]]]; 

    int\  = \ ∫u[t] \[DifferentialD]t\  + \ 
        I\ \(∫v[t] \[DifferentialD]t\); 

    val\  = \ ∫\_0\%1 u[t] \[DifferentialD]t\  + \ 
        I\ \(∫\_0\%1 v[t] \[DifferentialD]t\); 

    Print["                 f[t] = ", f[t]]; 

    Print["                 u[t] = ", u[t]]; 

    Print["                 v[t] = ", v[t]]; 

    Print["  ∫u[t]\[DifferentialD]t + I ∫v[t]\[DifferentialD]t = ", int]; 

    Print[\*""(∫_0^1)u[t]\[DifferentialD]t + I (∫_0^1)v[t]\[DifferentialD]t = "", val]; \)

Or we could do the integral directly, with complex function for the integrand.

\(int\  = \ ∫\(\((t - I)\)\^3\) \[DifferentialD]t; 
    
Print[\*"" ∫(t-I)^3\[DifferentialD]t = "", int]; 

    val\  = \ ∫\_0\%1 f[t] \[DifferentialD]t; 

    Print[\*""(∫_0^1)(t-I)^3\[DifferentialD]t = "", val]; \)

Example 6.2,  Page 158. Show that (∫_( 0)^( π/2)) 
            (e^(t  +  i t)) dt = (1/2)((e^((π/2)))-1)+(( i)/2)((e^((π/2)))+1).

\(Clear[f, int, t, u, v, val]; 

    f[t_]\  = \ ComplexExpand[E\^\(t + I\ t\)]; 

    u[t_]\  = \ ComplexExpand[Re[f[t]]]; 

    v[t_]\  = \ ComplexExpand[Im[f[t]]]; 

    int\  = \ ∫u[t] \[DifferentialD]t\  + \ 
        I\ \(∫v[t] \[DifferentialD]t\); 

    val\  = \ ∫\_0\%\(π/2\)u[t] \[DifferentialD]t\  + \ 
        I \(∫\_0\%\(π/2\)v[t] \[DifferentialD]t\); 
    
Print["               f[t] = ", f[t]]; 

    Print["               u[t] = ", u[t]]; 

    Print["               v[t] = ", v[t]]; 

    Print["∫u[t]\[DifferentialD]t + I ∫v[t]\[DifferentialD]t = ", int]; 

    Print[\*""(∫_0^(π/2))u[t]\[DifferentialD]t + I(∫_0^(π/2))v[t]\[DifferentialD]t = "", 
      val]; \)

Or we could do the integral directly, with complex function for the integrand.

\(\(\ int\  = \ 
      ∫\(E\^\(t + I\ t\)\) 
          \[DifferentialD]t; 

    Print[\*""   ∫\!\(E\^\(t + I\\ t\)\)\[DifferentialD]t = "", int]; 

    val\  = \ ∫\_0
          \%\(π/2\)\(E\^\(t + I\ t\)\) 
          \[DifferentialD]t; 

    Print[\*""(∫_0^(π/2))\!\(E\^\(t + I\\ t\)\)\[DifferentialD]t = "", val]; \)\)

Example 6.3,  Page 159.  Show that 
∫ (c + i d) (u(t) + i v(t))dt  =   c∫ u(t)dt - d∫ v(t)dt + i d∫ u(t)dt + i c ∫ v(t)dt

\(Clear[c, d, int1, int2, t, u, v]; 

    int1\  = ∫\((c\  + \ I\ d)\) 
          \((u[t]\  + \ I\ v[t])\) \[DifferentialD]t; 

    int2\  = \ MapAll[Distribute, int1]; 
Print[int1, " = "]\)Print[int2]

Example 6.4,  Page 159.  Show that   (∫_( 0)^( π)) 
            ((e )^(i t)) dt  =  2 i.

\(Clear[f, int, t, val]; 

    f[t_]\  = \ ComplexExpand[\(E\ \)\^\(I\ t\)]; 

    int\  = \ ∫f[t] \[DifferentialD]t; 

    val\  = \ ∫\_0\%π f[t] \[DifferentialD]t; 

    Print["    f[t] =   ", E\^\(I\ t\)\ , 
      "  = ", f[t]]; 

    Print[" ∫f[t]\[DifferentialD]t = ", 
      ∫\(E\^\(I\ t\)\) \[DifferentialD]t, 
      " = ", int]; 

    Print[\*""\!\(∫\_0\%π\)f[t]\[DifferentialD]t = "", 
      val]; \)

GoTo Chapter 6

Section 6.2  Contours and Contour Integrals

In this section we shall define and evaluate integrals of the form  (∫_C^( )) f(z) dz
where  f(z)  is complex-valued and  C  is a contour in the plane.

Theorem 6.1,  Page 166.  Suppose  f(z) is a continuous complex-valued function defined on a set containing the contour  C.  Let  z(t)  be any parameterization of  C  for  a ≤ t ≤ b.  Then
(∫_C^( )) f(z) dz=(∫_a^( b)) f(z(t)) z' (t) dt.

Example 6.5,  Page 162.  Let us find a parameterization of the polygonal path  C  
from  - 1 + i  to  3 - i  consisting of the three line segments:
Segment 1 from  - 1 + i  to  - 1,
Segment 2 from  - 1  to  1 + i,
Segment 3 from  1 + i  to  3 - i .

\(Clear[t, x0, x1, y0, y1, z1, Z, Z1]; 

    z0\  = \ \(-\ 1\)\  + \ I; 
z1\  = \ \(-\ 1\); 

    x0\  = \ Re[z0]; y0\  = \ Im[z0]; x1\  = \ Re[z1]; y1\  = \ Im[z1]; 

    Z1[t_]\  = \ 
      x0\  + \ \((x1 - x0)\) t\  + \ 
        I \((y0\  + \ \((y1 - y0)\) t)\); 
z0\  = \ \(-\ 1\); 

    z1\  = \ 1\  + \ I; 
x0\  = \ Re[z0]; y0\  = \ Im[z0]; 
    x1\  = \ Re[z1]; y1\  = \ Im[z1]; 

    Z2[t_]\  = \ 
      x0\  + \ \((x1 - x0)\) t\  + \ 
        I \((y0\  + \ \((y1 - y0)\) t)\); 

    z0\  = \ 1\  + \ I; 
z1\  = \ 3\  - \ I; 

    x0\  = \ Re[z0]; y0\  = \ Im[z0]; x1\  = \ Re[z1]; y1\  = \ Im[z1]; 

    Z3[t_]\  = \ 
      x0\  + \ \((x1 - x0)\) t\  + \ 
        I \((y0\  + \ \((y1 - y0)\) t)\); 

    Print[\*""\!\(Z\_1\)[t] = "", Z1[t]]; 

    Print[\*""\!\(Z\_2\)[t] = "", Z2[t]]; 

    Print[\*""\!\(Z\_3\)[t] = "", Z3[t]]; \)

Example 6.6,  Page 164.  Use a Riemann sum to construct an approximation 
for the contour integral   (∫_C^( ))  (e^z)dz   where  C  is a the line segment joining 
the point   a = 0   to   b = 2 + ((i π)/4).

\(Clear[a, b, c, dz, f, k, n, RiemannSum, z]; 

    f[z_]\  = \ E\^z; 
a\  = \ 0; 

    b\  = \ 2\  + \(\ I\ π\)\/4; 
n\  = \ 8; 

    dz\  = \ \(b - a\)\/n; 
z[k_]\  = \ a\  + \ k\ dz; 

    c[k_]\  = \ Together[\(z[k - 1] + \ z[k]\)\/2]; 

    RiemannSum\  = ∑\+\(k = 1\)\%8 f[c[k]] dz\ ; 

    Print["f[z] = ", f[z]]; 

    Print[\*""\!\(∑\+\(k = 1\)\%8\)f[\!\(c\_k\)]\!\(dz\_k\) = ""]; 
    
Print[RiemannSum]; 

    Print[\*""\!\(∑\+\(k = 1\)\%8\)f[\!\(c\_k\)]\!\(dz\_k\) = ""]; 
    
Print[\((∑\+\(k = 1\)\%8 f[c[k]])\) dz]; 

    Print[\*""\!\(∑\+\(k = 1\)\%8\)f[\!\(c\_k\)]\!\(dz\_k\) = "", 
      N[RiemannSum]]; \)

Example 6.7,  Page 166.  Evaluate the contour integral   (∫_C^( )) exp(z) dz   
where  C  is a the line segment joining the point   a = 0   to   b = 2 + ((i π)/4).

\(Clear[a, b, f, int, t, u, v, val, z, Z]; 
a\  = \ 0; 

    b\  = \ 2\  + \(\ I\ π\)\/4; 

    z[t_]\  = \ \ a\  + \ \((b\  - \ a)\)\ t; 

    F[Z_]\  = \ E\^Z; 
f[t_]\  = \ E\^z[t]; 

    u[t_]\  = \ ComplexExpand[Re[f[t]]]; 

    v[t_]\  = \ ComplexExpand[Im[f[t]]]; 
Print["     F[z] = ", F[z]]; 

    Print["     z[t] = ", z[t]]; 

    Print["     f[t] = F[z[t]] = ", f[t]]; 

    Print["     u[t] = ", u[t]]; 
Print["     v[t] = ", v[t]]; 

    Print["    z'[t] = ", \(z'\)[t]]; 

    Print["f[t]z'[t] = ", f[t] \(z'\)[t]]; \)

\(int\  = ∫f[t] \(z'\)[t] \[DifferentialD]t; 

    val\  = \ ∫\_0\%1 f[t] \(z'\)[t] \[DifferentialD]t; 

    Print[" ∫f[t]z'[t]\[DifferentialD]t = ", int]; 

    Print[\*""\!\(∫\_0\%1\)f[t]z'[t]\[DifferentialD]t = "", 
      val]; 
Print[
      \*""\!\(∫\_0\%1\)f[t]z'[t]\[DifferentialD]t = "", 
      ComplexExpand[val]]; 

    Print[\*""\!\(∫\_0\%1\)f[t]z'[t]\[DifferentialD]t = "", 
      N[ComplexExpand[val]]]; \)

\(int\  = \ 
      ∫u[t] \(z'\)[t] \[DifferentialD]t\  + \ 
        I\ \(∫v[t] \(z'\)[t] \[DifferentialD]t\); 

    val\  = \ \ 
      ∫\_0\%1 u[t] \(z'\)[t] \[DifferentialD]t\  + \ 
        I\ 
          \(∫\_0\%1 v[t] \(z'\)[t] \[DifferentialD]t\); 

    Print["∫u[t]z'[t]\[DifferentialD]t + I ∫v[t]z'[t]\[DifferentialD]t"]; 
Print["= ", int]; 

    Print[" "]; 

    Print[\*""(∫_0^1)u[t]z'[t]\[DifferentialD]t + I (∫_0^1)v[t]z'[t]\[DifferentialD]t""]; 

    Print["= ", val]; 
Print["= ", ComplexExpand[val]]; 

    Print["= ", Apart[ComplexExpand[val]]]; 

    Print["= ", N[ComplexExpand[val]]]; \)

Example 6.8,  Page 167.  Evaluate the contour integral   (∫_C^( )) (1/(z - 2))dz
where  C  is a the upper semicircle with radius  1  centered at  x = 2.

\(Clear[f, int, t, val, z]; 
f[z_]\  = \ 1\/\(z\  - \ 2\); 

    z[t_]\  = \ 2\  + \ E\^\(I\ t\); 

    int\  = \ ∫f[z[t]] \(z'\)[t] \[DifferentialD]t; 

    val\  = \ ∫\_0\%π f[z[t]] \(z'\)[t] \[DifferentialD]t; 

    Print["               f[z] = ", f[z]]; 

    Print["               z[t] = ", z[t]]; 

    Print["            f[z[t]] = ", f[z[t]]]; 

    Print["              z'[t] = ", \(z'\)[t]]; 

    Print["       f[z[t]]z'[t] = ", f[z[t]] \(z'\)[t]]; 

    Print["    ∫f[z[t]]z'[t]\[DifferentialD]t = ", int]; 

    Print[\*""   (∫_0^π)f[z[t]]z'[t]\[DifferentialD]t = "", val]; \)

Example 6.9,  Page 168.  Evaluate the contour integrals  of   f(z) = z   over a line segment and portion of a parabola.

\(Clear[f, t, u, U, U1, U2, v, V, V1, V2, w, W, W1, W2, z, z1, z2]; 

    f[z_]\  = \ z; 
Print["f[z] = ", f[z]]; \)

(a)  Use the straight line connecting the points   -1 - i   and   3 + i.

\(z1[t_]\  = \ 2  t\  + \ 1\  + \ I\ t; 

    w[t_]\  = \ ComplexExpand[f[z1[t]]\ \(z1'\)[t]]; 

    u[t_]\  = \ ComplexExpand[Re[w[t]]]; 

    v[t_]\  = \ ComplexExpand[Im[w[t]]]; 

    Print[\*""          \!\(z\_1\)[t] = "", z1[t]]; 

    Print[\*""       f[\!\(z\_1\)[t]] = "", f[z1[t]]]; 

    Print[\*""         \!\(z\_1\)'[t] = "", \(z1'\)[t]]; 

    Print[\*""f[\!\(z\_1\)[t]] \!\(z\_1\)'[t] = "", w[t]]; 

    Print["           u[t] = ", u[t]]; 

    Print["           v[t] = ", v[t]]; \)

\(U1\  = ∫\_\(-1\)\%1 u[t] \[DifferentialD]t; 

    V1\  = ∫\_\(-1\)\%1 v[t] \[DifferentialD]t; 

    W1\  = \ U1\  + \ I\ V1; 

    U[t_]\  = \ ∫u[t] \[DifferentialD]t; 

    V[t_]\  = \ ∫v[t] \[DifferentialD]t; 

    int\  = \ ∫u[t] \[DifferentialD]t\  + \ 
        I\ \(∫v[t] \[DifferentialD]t\); 

    Print["∫u[t]\[DifferentialD]t + I ∫v[t]\[DifferentialD]t = ", int]; 

    Print["U[1] = ", U[1], ",      V[1] = ", V[1]]; 

    Print["U[-1] = ", U[\(-1\)], ",   V[-1] = ", V[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)u[t]\[DifferentialD]t + I(∫_(-1)^1)v[t]\[DifferentialD]t = U[1]-U[-1] + I(V[1]-V[-1])""]; 

    Print[\*""(∫_(-1)^1)u[t]\[DifferentialD]t + I(∫_(-1)^1)v[t]\[DifferentialD]t = "", W1]; 
    \)

\(W[t_]\  = ∫w[t] \[DifferentialD]t; 

    W1\  = ∫\_\(-1\)\%1 w[t] \[DifferentialD]t; 

    Print["W[t] = ∫w[t]\[DifferentialD]t = ", W[t]]; 

    Print["W[1] = ", W[1], ",  W[-1] = ", W[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)w[t]\[DifferentialD]t = W[1]-W[-1] = "", W[1] - W[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)w[t]\[DifferentialD]t = "", 
      W1]; \)

(b)  Use the portion of a parabola connecting the points   -1 - i   and   3 + i.

\(z2[t_]\  = \ t\^2\  + \ 2  t\  + \ I\ t; 

    w[t_]\  = \ ComplexExpand[f[z2[t]]\ \(z2'\)[t]]; 

    u[t_]\  = \ ComplexExpand[Re[w[t]]]; 

    v[t_]\  = \ ComplexExpand[Im[w[t]]]; 

    Print[\*""          \!\(z\_2\)[t] = "", z2[t]]; 

    Print[\*""       f[\!\(z\_2\)[t]] = "", f[z2[t]]]; 

    Print[\*""         \!\(z\_2\)'[t] = "", \(z2'\)[t]]; 

    Print[\*""f[\!\(z\_2\)[t]] \!\(z\_2\)'[t] = "", w[t]]; 

    Print["           u[t] = ", u[t]]; 

    Print["           v[t] = ", v[t]]; \)

\(U2\  = ∫\_\(-1\)\%1 u[t] \[DifferentialD]t; 

    V2\  = ∫\_\(-1\)\%1 v[t] \[DifferentialD]t; 

    W2\  = \ U2\  + \ I\ V2; 

    U[t_]\  = \ ∫u[t] \[DifferentialD]t; 

    V[t_]\  = \ ∫v[t] \[DifferentialD]t; 

    int\  = \ ∫u[t] \[DifferentialD]t\  + \ 
        I\ \(∫v[t] \[DifferentialD]t\); 

    Print["∫u[t]\[DifferentialD]t + I ∫v[t]\[DifferentialD]t = ", int]; 

    Print["U[1] = ", U[1], ",     V[1] = ", V[1]]; 

    Print["U[-1] = ", U[\(-1\)], ",   V[-1] = ", V[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)u[t]\[DifferentialD]t + I(∫_(-1)^1)v[t]\[DifferentialD]t = U[1]-U[-1] + I(V[1]-V[-1])""]; 

    Print[\*""(∫_(-1)^1)u[t]\[DifferentialD]t + I(∫_(-1)^1)v[t]\[DifferentialD]t = "", W2]; 
    \)

\(W[t_]\  = ∫w[t] \[DifferentialD]t; 

    W2\  = ∫\_\(-1\)\%1 w[t] \[DifferentialD]t; 

    Print["W[t] = ∫w[t]\[DifferentialD]t = ", W[t]]; 

    Print["W[1] = ", W[1], ",  W[-1] = ", W[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)w[t]\[DifferentialD]t = W[1]-W[-1] = "", W[1] - W[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)w[t]\[DifferentialD]t = "", 
      W2]; \)

Which is the same as the value for the other path.

\(\(Print[
      \*""\!\(∫\_\(-1\)\%1\)w[t]\[DifferentialD]t = "", W1]; 
    \)\)

Example 6.* after,  Page 168.  Similar to the above example, but using \(TraditionalForm\`z\&_\).  Evaluate the contour integrals of   f(z) = \(TraditionalForm\`z\&_\)  over a line segment and portion of a parabola.

\(Clear[f, t, u, U, U1, U2, v, V, V1, V2, w, W, W1, W2, z, z1, z2]; 

    f[z_]\  = \ Conjugate[z]; 
Print["   f[z] = ", f[z]]; 

    Print["f[x+Iy] = ", 
      ComplexExpand[f[x + I\ y]]]; \)

(a)  Use the straight line connecting the points   -1 - i   and   3 + i.

\(t /: Im[t] = 0; 
z1[t_]\  = \ 2  t\  + \ 1\  + \ I\ t; 

    w[t_]\  = \ ComplexExpand[f[z1[t]]\ \(z1'\)[t]]; 

    u[t_]\  = \ ComplexExpand[Re[w[t]]]; 

    v[t_]\  = \ ComplexExpand[Im[w[t]]]; 

    Print[\*""          \!\(z\_1\)[t] = "", z1[t]]; 

    Print[\*""       f[\!\(z\_1\)[t]] = "", 
      ComplexExpand[\ f[z1[t]]\ ]]; 

    Print[\*""         \!\(z\_1\)'[t] = "", \(z1'\)[t]]; 

    Print[\*""f[\!\(z\_1\)[t]] \!\(z\_1\)'[t] = "", w[t]]; 

    Print["           u[t] = ", u[t]]; 

    Print["           v[t] = ", v[t]]; \)

\(U1\  = ∫\_\(-1\)\%1 u[t] \[DifferentialD]t; 

    V1\  = ∫\_\(-1\)\%1 v[t] \[DifferentialD]t; 

    W1\  = \ U1\  + \ I\ V1; 

    U[t_]\  = \ ∫u[t] \[DifferentialD]t; 

    V[t_]\  = \ ∫v[t] \[DifferentialD]t; 

    int\  = \ ∫u[t] \[DifferentialD]t\  + \ 
        I\ \(∫v[t] \[DifferentialD]t\); 

    Print["∫u[t]\[DifferentialD]t + I ∫v[t]\[DifferentialD]t = ", int]; 

    Print["U[1] = ", U[1], ",     V[1] = ", V[1]]; 

    Print["U[-1] = ", U[\(-1\)], ",   V[-1] = ", V[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)u[t]\[DifferentialD]t + I(∫_(-1)^1)v[t]\[DifferentialD]t = U[1]-U[-1] + I(V[1]-V[-1])""]; 

    Print[\*""(∫_(-1)^1)u[t]\[DifferentialD]t + I (∫_(-1)^1)v[t]\[DifferentialD]t = "", W1]; 
    \)

\(W[t_]\  = ∫w[t] \[DifferentialD]t; 

    W1\  = ∫\_\(-1\)\%1 w[t] \[DifferentialD]t; 

    Print["W[t] = ∫w[t]\[DifferentialD]t = ", W[t]]; 

    Print["W[1] = ", W[1], ",  W[-1] = ", W[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)w[t]\[DifferentialD]t = W[1]-W[-1] = "", W[1] - W[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)w[t]\[DifferentialD]t = "", 
      W1]; \)

(b)  Use the portion of a parabola connecting the points   -1 - i   and   3 + i.

\(t /: Im[t] = 0; 
z2[t_]\  = \ t\^2\  + \ 2  t\  + \ I\ t; 
    
w[t_]\  = \ ComplexExpand[f[z2[t]]\ \(z2'\)[t]]; 

    u[t_]\  = \ ComplexExpand[Re[w[t]]]; 

    v[t_]\  = \ ComplexExpand[Im[w[t]]]; 

    Print[\*""          \!\(z\_2\)[t] = "", z2[t]]; 

    Print[\*""       f[\!\(z\_2\)[t]] = "", 
      ComplexExpand[\ f[z2[t]]\ ]]; 

    Print[\*""         \!\(z\_2\)'[t] = "", \(z2'\)[t]]; 

    Print[\*""f[\!\(z\_2\)[t]] \!\(z\_2\)'[t] = "", w[t]]; 

    Print["           u[t] = ", u[t]]; 

    Print["           v[t] = ", v[t]]; \)

\(U2\  = ∫\_\(-1\)\%1 u[t] \[DifferentialD]t; 

    V2\  = ∫\_\(-1\)\%1 v[t] \[DifferentialD]t; 

    W2\  = \ U2\  + \ I\ V2; 

    U[t_]\  = \ ∫u[t] \[DifferentialD]t; 

    V[t_]\  = \ ∫v[t] \[DifferentialD]t; 

    int\  = \ ∫u[t] \[DifferentialD]t\  + \ 
        I\ \(∫v[t] \[DifferentialD]t\); 

    Print["∫u[t]\[DifferentialD]t + I ∫v[t]\[DifferentialD]t = ", int]; 

    Print["U[1] = ", U[1], ",     V[1] = ", V[1]]; 

    Print["U[-1] = ", U[\(-1\)], ",   V[-1] = ", V[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)u[t]\[DifferentialD]t + I(∫_(-1)^1)v[t]\[DifferentialD]t = U[1]-U[-1] + I(V[1]-V[-1])""]; 

    Print[\*""(∫_(-1)^1)u[t]\[DifferentialD]t + I (∫_(-1)^1)v[t]\[DifferentialD]t = "", W2]; 
    \)

\(W[t_]\  = ∫w[t] \[DifferentialD]t; 

    W2\  = ∫\_\(-1\)\%1 w[t] \[DifferentialD]t; 

    Print["W[t] = ∫w[t]\[DifferentialD]t = ", W[t]]; 

    Print["W[1] = ", W[1], ",  W[-1] = ", W[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)w[t]\[DifferentialD]t = W[1]-W[-1] = "", W[1] - W[\(-1\)]]; 

    Print[\*""(∫_(-1)^1)w[t]\[DifferentialD]t = "", 
      W2]; \)

Which is different from the value for the other path.

\(\(Print[
      \*""\!\(∫\_\(-1\)\%1\)w[t]\[DifferentialD]t = "", W1]; 
    \)\)

Example 6.10,  Page 169.  (a)   Show that   (∫_(Subscript[C, 1])^( )) \(z\&_\ \ dz\)  =  - π i
where (Subscript[C, 1]) is the semicircular path from  -1  to  1  in the upper half plane.

(b)  Show that   (∫_(Subscript[C, 2])^( )) \(z\&_\ \ dz\)   =  - 4 i 
where (Subscript[C, 2]) is the polygonal path from   -1   to   -1+ i   to   1 + i   to 1.

Clear[f, t, u, v, w, W, W1, z]\(f[z_]\  = \ Conjugate[z]; 
Print["   f[z] = ", f[z]]; 

    Print["f[x+Iy] = ", 
      ComplexExpand[f[x + I\ y]]]; \)

(a)  Use the semicircular path from  -1  to  1  in the upper half plane.

\(t /: Im[t] = 0; 

    z[t_]\  = \ \ \(-\ Cos[t]\)\  + \ I\ Sin[t]; 

    w[t_]\  = \ ComplexExpand[f[z[t]]\ \(z'\)[t]]; 

    u[t_]\  = \ ComplexExpand[Re[w[t]]]; 

    v[t_]\  = \ ComplexExpand[Im[w[t]]]; 

    Print["         z[t] = ", z[t]]\)Print["      f[z[t]] = ", ComplexExpand[f[z[t]]]]Print["        z'[t] = ", z'[t]]Print["f[z[t]] z'[t] = ", w[t]]\(w[t_]\  = \ Simplify[w[t]]; 
Print["f[z[t]] z'[t] = ", w[t]]\)Print["          u[t] = ", u[t]]Print["          v[t] = ", v[t]]\(v[t_]\  = \ Simplify[v[t]]; 
Print["          v[t] = ", v[t]]\)

\(W[t_]\  = ∫w[t] \[DifferentialD]t; 

    W1\  = ∫\_0\%π w[t] \[DifferentialD]t; 

    Print["W[t] = ∫w[t]\[DifferentialD]t = ", W[t]]; 

    Print["W[π] = ", W[π], ",  W[0] = ", W[0]]; 

    Print[\*""\!\(∫\_\(\\ 0\)\%π\)w[t]\[DifferentialD]t = W[π]-W[0] = "", W[π] - W[0]]; 

    Print[\*""\!\(∫\_\(\\ 0\)\%π\)w[t]\[DifferentialD]t = "", W1]; \)

(b)  Use the  polygonal path from  - 1 to  - 1 + i  to 1 + i  to 1.  Let us find a parameterization of the polygonal path  C  from  - 1 to  1  consisting of the three line segments:
Segment 1 from   - 1   to   - 1 + i ,
Segment 2 from   - 1 + i   to   1 + i ,
Segment 3 from   1 + i   to 1.

\(Clear[t, x0, x1, y0, y1, z1, Z, Z1]; 
z0\  = \ \(-\ 1\); 

    z1\  = \ \(-\ 1\)\  + \ I; 
x0\  = \ Re[z0]; 
    y0\  = \ Im[z0]; x1\  = \ Re[z1]; y1\  = \ Im[z1]; 

    Z1[t_]\  = \ 
      x0\  + \ \((x1 - x0)\) t\  + \ 
        I \((y0\  + \ \((y1 - y0)\) t)\); 

    z0\  = \ \(-\ 1\)\  + \ I; 
z1\  = \ 1\  + \ I; 
    
x0\  = \ Re[z0]; y0\  = \ Im[z0]; x1\  = \ Re[z1]; y1\  = \ Im[z1]; 

    Z2[t_]\  = \ 
      x0\  + \ \((x1 - x0)\) t\  + \ 
        I \((y0\  + \ \((y1 - y0)\) t)\); 

    z0\  = \ 1\  + \ I; 
z1\  = \ 1; 
x0\  = \ Re[z0]; 
    y0\  = \ Im[z0]; x1\  = \ Re[z1]; y1\  = \ Im[z1]; 

    Z3[t_]\  = \ 
      x0\  + \ \((x1 - x0)\) t\  + \ 
        I \((y0\  + \ \((y1 - y0)\) t)\); 

    Print[\*""\!\(Z\_1\)[t] = "", Z1[t]]; 

    Print[\*""\!\(Z\_2\)[t] = "", Z2[t]]; 

    Print[\*""\!\(Z\_3\)[t] = "", Z3[t]]; \)

\(Wset\  = 
      \ {f[Z1[t]] \(Z1'\)[t], f[Z2[t]] \(Z2'\)[t], f[Z3[t]] \(Z3'\)[t]}; 

    funset\  = \ MapAll[ComplexExpand, Wset]; 

    Wint\  = ∫funset \[DifferentialD]t; 

    val\  = ∫\_0\%1 funset \[DifferentialD]t; 

    val2\  = ReplaceAll[val\ , List -> Plus]; 

    Print[\*""{(Subscript[w, 1])[t],(Subscript[w, 2])[t],(Subscript[w, 3])[t]}""]\ ; 

    Print[\*""= {f[(Subscript[Z, 1])[t]](Subscript[Z, 1])'[t],f[(Subscript[Z, 2])[t]](Subscript[Z, 2])'[t],f[(Subscript[Z, 3])[t]](Subscript[Z, 3])'[t]}""]; 
Print[" "]; 
Print["= ", funset]; 

    Print[" "]; 

    Print[\*""∫{(Subscript[w, 1])[t],(Subscript[w, 2])[t],(Subscript[w, 3])[t]}\[DifferentialD]t = ""\ , Wint]; 
Print[" "]; 

    Print[\*""(∫_0^1){(Subscript[w, 1])[t],(Subscript[w, 2])[t],(Subscript[w, 3])[t]}\[DifferentialD]t = "", val]; 
Print[" "]; 

    Print[\*""(∫_0^1)w[t]\[DifferentialD]t = "", val2]; 
    \)

Which is different from the value for the other path.

\(\(Print[
      \*""\!\(∫\_\(\\ 0\)\%π\)w[t]\[DifferentialD]t = "", W1]; \)\)

Example 6.11,  Page 173.    Use the ML Inequality to show that
Show that   |(∫_C^( )) (1/((z^2)  +  1)) dz|≤(1/(2 (Sqrt[5]))).
Solution.  The solution is given in the textbook.

GoTo Chapter 6

Section 6.3  The Cauchy-Goursat Theorem

The Cauchy-Goursat theorem states that within certain domains the 
integral of an analytic function over a simple closed contour is zero.

Theorem 6.2,  Page 177.  (Green's Theorem)  Let  C  be a simple closed contour with positive orientation, and let  R  be the domain that forms the interior of  C.  If  P  and  Q  are continuous and have continuous partial derivatives at all points on  C  and  R, then  (∫_C^( )) P(x, y) dx+(∫_C^( )) Q(x, y) dy=∫∫(Underscript[  , R]) [(∂_x) Q(x, y) - (∂_y) P(x, y)]
                   dx dy.

Theorem 6.3,  Page 179.  (Cauchy-Goursat Theorem)  Let  f(z)  be analytic in a simply connected domain  D.  If  C  is a simple closed contour that lies in  D,  then  (∫_C^( )) f(z) dz=0.

Theorem 6.4,  Page 183.  (Deformation of Contour)  Let  (Subscript[C, 1])  and  (Subscript[C, 2])  be two simple closed positively oriented contours such that  (Subscript[C, 1])  lies interior to (Subscript[C, 2]).  If   f(z)  is analytic in a domain  D  that contains both  (Subscript[C, 1])  and (Subscript[C, 2])  and the region between them, then  (∫_(Subscript[C, 1])^( )) f(z) dz=(∫_(Subscript[C, 2])^( )) f(z) dz.

Theorem 6.5,  Page 186.  (Extended Cauchy-Goursat Theorem)  Let  C, (Subscript[C, 1]), (Subscript[C, 2]), ⋯, (Subscript[C, n])  be simple closed positively oriented contours with the property that  (Subscript[C, k])  lies interior to  C  for   k = 1, 2, ⋯, n  and the set of interior to  (Subscript[C, k])  has no points in common with the set interior to  (Subscript[C, j])  if  k ≠ j.  Let  f(z)  be analytic on a domain  D  that contains all the contours and the region between  C  and  (Subscript[C, 1]) + (Subscript[C, 2]) + ⋯ + (Subscript[C, n]), then  (∫_C^( )) f(z) dz=(∑_(k = 1)^n)(∫_(Subscript[C, k])^( )) f(z) dz.

Example 6.12,  Page 182.  Let us recall that  (e^z),  cos(z),  and  (z^n),  where n is a positive integer, are all entire functions and have continuous derivatives.  For illustration set  n=5, and choose the contour to be the unit circle C: |z| = 1 with positive orientation.  The Cauchy-Goursat theorem implies that for any simple closed contour we have:

(∫_C^( )) exp(z) dz  =  0,

(∫_C^( )) cos(z) dz  =  0  and

(∫_C^( )) (z^5) dz  =  0.

First, use the function  f(z) = (e^z).

\(Clear[f, F, θ, z]; 
f[z_]\  = \ E\^z; 

    z[θ_]\  = \ E\^\(I\ θ\); 

    F[θ_]\  = 
      ∫f[z[θ]] \(z'\)[θ] \[DifferentialD]θ; 

    val\  = \ ∫\_0\%\(2  π\)f[z[θ]] \(z'\)[θ] 
          \[DifferentialD]θ; 

    Print["              f[z] = ", f[z]]; 

    Print["              z[θ] = ", z[θ]]; 

    Print["           f[z[θ]] = ", f[z[θ]]]; 

    Print["             z'[θ] = ", \(z'\)[θ]]; 

    Print["     f[z[θ]] z'[θ] = ", 
      f[z[θ]]\ \(z'\)[θ]]; 

    Print["   ∫f[z[θ]]z'[θ]\[DifferentialD]t = ", 
      F[θ]]; 

    Print["F[2π] = ", F[2  π], ",    F[0] = ", F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = F[2π]-F[0] = "", 
F[2  π] - F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = "", val]; \)

Second, use the function  f(z) = cos(z).

Clear[f, F, θ, z]\(f[z_]\  = \ Cos[z]; 

    z[θ_]\  = \ E\^\(I\ θ\); 

    F[θ_]\  = 
      ∫f[z[θ]] \(z'\)[θ] \[DifferentialD]θ; 

    val\  = \ ∫\_0\%\(2  π\)f[z[θ]] \(z'\)[θ] 
          \[DifferentialD]θ; 

    Print["              f[z] = ", f[z]]; 

    Print["              z[θ] = ", z[θ]]; 

    Print["           f[z[θ]] = ", f[z[θ]]]; 

    Print["             z'[θ] = ", \(z'\)[θ]]; 

    Print["     f[z[θ]] z'[θ] = ", 
      f[z[θ]]\ \(z'\)[θ]]; 

    Print["   ∫f[z[θ]]z'[θ]\[DifferentialD]t = ", 
      F[θ]\ ]; 

    Print["F[2π] = ", F[2  π], ",    F[0] = ", F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = F[2π]-F[0] = "", 
F[2  π] - F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = "", val]; \)

Third, use the function  f(z) = (z^5).

Clear[f, F, θ, z]\(f[z_]\  = \ z\^5; 

    z[θ_]\  = \ E\^\(I\ θ\); 

    F[θ_]\  = 
      ∫f[z[θ]] \(z'\)[θ] \[DifferentialD]θ; 

    val\  = \ ∫\_0\%\(2  π\)f[z[θ]] \(z'\)[θ] 
          \[DifferentialD]θ; 

    Print["              f[z] = ", f[z]]; 

    Print["              z[θ] = ", z[θ]]; 

    Print["           f[z[θ]] = ", f[z[θ]]]; 

    Print["             z'[θ] = ", \(z'\)[θ]]; 

    Print["     f[z[θ]] z'[θ] = ", 
      f[z[θ]]\ \(z'\)[θ]]; 

    Print["   ∫f[z[θ]]z'[θ]\[DifferentialD]t = ", 
      F[θ]]; 

    Print["F[2π] = ", F[2  π], ",    F[0] = ", F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = F[2π]-F[0] = "", 
F[2  π] - F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = "", val]; \)

Example 6.13,  Page 182.  If  C  is a simple closed contour such that the origin does not lie interior to  C,  then there is a simply connected domain  D  that contains  C  in which  f(z)=(1/((z^n)))  is analytic.  The Cauchy-Goursat theorem implies that  (∫_C^( )) (1/((z^n)))dz  =  0,  provided that the origin does not lie interior to  C.
Solution.  The solution is given in the textbook.

Example 6.14,  Page 184.   Let  (Subscript[z, 0])  denote a fixed complex value.   If  C  is a 
simple closed contour with positive orientation such that  (Subscript[z, 0])  lies interior to C, then

(∫_C^( )) (1/(z - (Subscript[z, 0])))dz  =  2 π i ,

(∫_C^( )) (1/(((z - (Subscript[z, 0])))^n))dz  =  0 , where  n ≠ 1 is an integer.

(a)   Show that   (∫_C^( )) (1/(z - (Subscript[z, 0])))dz  =  2 π i .
For illustration, we use a circle of radius 1 centered at  (Subscript[z, 0]).

\(Clear[f, F, θ, z, z0]; 
f[z_]\  = \ 1\/\(z - z0\); 

    z[θ_]\  = \ z0\  + \ E\^\(I\ θ\); 
    
F[θ_]\  = 
      ∫f[z[θ]] \(z'\)[θ] \[DifferentialD]θ; 

    val\  = \ ∫\_0\%\(2  π\)f[z[θ]] \(z'\)[θ] 
          \[DifferentialD]θ; 

    Print["              f[z] = ", f[z]]; 

    Print["              z[θ] = ", z[θ]]; 

    Print["           f[z[θ]] = ", f[z[θ]]]; 

    Print["             z'[θ] = ", \(z'\)[θ]]; 

    Print["     f[z[θ]] z'[θ] = ", 
      f[z[θ]]\ \(z'\)[θ]]; 

    Print["   ∫f[z[θ]]z'[θ]\[DifferentialD]t = ", 
      F[θ]]; 

    Print["F[2π] = ", F[2  π], ",    F[0] = ", F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = F[2π]-F[0] = "", 
F[2  π] - F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = "", val]; \)

(b)   Show that   (∫_C^( )) (1/(((z - (Subscript[z, 0])))^5))dz  =  0 ,
For illustration, we use a circle of radius 1 centered at  (Subscript[z, 0]).

Clear[f, F, θ, z, z0]\(f[z_]\  = \ 1\/\((z - z0)\)\^5; 

    z[θ_]\  = \ z0\  + \ E\^\(I\ θ\); 
    
F[θ_]\  = 
      ∫f[z[θ]] \(z'\)[θ] \[DifferentialD]θ; 

    val\  = \ ∫\_0\%\(2  π\)f[z[θ]] \(z'\)[θ] 
          \[DifferentialD]θ; 

    Print["              f[z] = ", f[z]]; 

    Print["              z[θ] = ", z[θ]]; 

    Print["           f[z[θ]] = ", f[z[θ]]]; 

    Print["             z'[θ] = ", \(z'\)[θ]]; 

    Print["     f[z[θ]] z'[θ] = ", 
      f[z[θ]]\ \(z'\)[θ]]; 

    Print["   ∫f[z[θ]]z'[θ]\[DifferentialD]t = ", 
      F[θ]]; 

    Print["F[2π] = ", F[2  π], ",    F[0] = ", F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = F[2π]-F[0] = "", 
F[2  π] - F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = "", val]; \)

Example 6.15,  Page 186.  Show that   (∫_C^( )) ((2 z)/((z^2)  +  2))dz  =  4 π i ,
where  C  is the circle  C: |z| = 2  taken with positive orientation.

\(Clear[f, F, i, p, R, solset, z, Z]; 

    f[z_]\  = \ \(2  z\)\/\(z\^2\  + \ 2\); 
Print["f[z] = ", f[z]]; \)

First, split the function up into partial fractions involving linear terms in the denominators. Don't worry about the subroutine.  It splits f(z) up into partial fractions.  The result will be an equivalent function named  F[z].  The list of terms added to form F[z] have singularities in the same order in the list of points  Z.

\(solset\  = \ Sort[Solve[Denominator[f[z]] == 0, z]]; 

    Z\  = \ Table[0, {Length[solset]}]; 
R\  = \ Table[0, {Length[solset]}]; 
    
Clear[F]; \ F[z_]\  = \ 0; 

    For[i = 1, i ≤ Length[solset], \(i++\), 
\ \ \ 
      Z[[i]]\  = \ 
        ReplaceAll[z, solset[[i]]]; \ 

      \ \ \ If[1 == i, p\  = \ 1]; 
\ \ \ 
      If[1 < i; 
        Z[[i - 1]] == 
          Z[[i]], p\  = \ p + 1, 
        p\  = \ 1]; 
\ \ \ 
      R[[i]]\  = \ 
        Residue[\((z - Z[[i]])\)\^\(p - 1
                \)\ f[z], {z, Z[[i]]}]; 
      
\ \ \ F[z_]\  = \ 
        F[z]\  + \ 
          R[[i
              ]]\/\((
                z - Z[[i]])\)\^p; ]; 

    For[i = 1, i ≤ Length[F[z]], \(i++\), 
\ \ \ 
      soset\  = \ 
        Solve[Denominator[\(F[z]\)[[i]]]
               == 0, z]; 
\ \ \ 
      Z[[i]]\  = \ 
        ReplaceAll[z, soset[[1]]]; 

      \ \ \ R[[i]]\  = \ 
        Residue[f[z], {z, Z[[i]]}]; ]; 

    Print["f[z] = ", f[z]]; 
Print["f[z] = ", F[z]]; \)

Determine which singularities lie inside C:   |z| = 2.

\(Print[\*""\!\(z\_1\) = "", 
      Z[[1]], 
",    Is |", 
      Z[[1]], "| < 2 ?    ", 
      N[Abs[Z[[1]]]] < 2]; 

    Print[\*""\!\(z\_2\) = "", 
      Z[[2]], 
",  Is |", 
      Z[[2]], "| < 2 ?   ", 
      N[Abs[Z[[2]]]] < 2]; \)

Since both singularities lie inside C, add the corresponding term in the sum forming  F[z]
times constant in the numerator that term times  2 π i.

\(val\  = \ 
      2  π\ I\ 
          Numerator[\(F[z]\)[[1]]] + 
        2  π\ I\ 
          Numerator[\(F[z]\)[[2]]]; 

    Print[\*""\!\(\(∫\_C\%.\)\)\!\(\(2  z\)\/\(z\^2\\  + \\ 2\)\)\[DifferentialD]z = "", val]; \)

Example 6.16,  Page 187.  Show that   (∫_C^( )) ((2 z)/((z^2)  +  2))dz  =  2 π i ,
where  C  is the circle  C: |z - i| = 1  taken with positive orientation.

\(Clear[f, F, i, p, R, solset, z, Z]; 

    f[z_]\  = \ \(2  z\)\/\(z\^2\  + \ 2\); 
Print["f[z] = ", f[z]]; \)

First, split the function up into partial fractions involving linear terms in the denominators.  Don't worry about the subroutine. It splits f(z) up into partial fractions.  The result will be an equivalent function named  F[z].  The list of terms added to form F[z] have singularities in the same order in the list of points  Z.

\(solset\  = \ Sort[Solve[Denominator[f[z]] == 0, z]]; 

    Z\  = \ Table[0, {Length[solset]}]; 
R\  = \ Table[0, {Length[solset]}]; 
    
Clear[F]; \ F[z_]\  = \ 0; 

    For[i = 1, i ≤ Length[solset], \(i++\), 
\ \ \ 
      Z[[i]]\  = \ 
        ReplaceAll[z, solset[[i]]]; \ 

      \ \ \ If[1 == i, p\  = \ 1]; 
\ \ \ 
      If[1 < i; 
        Z[[i - 1]] == 
          Z[[i]], p\  = \ p + 1, 
        p\  = \ 1]; 
\ \ \ 
      R[[i]]\  = \ 
        Residue[\((z - Z[[i]])\)\^\(p - 1
                \)\ f[z], {z, Z[[i]]}]; 
      
\ \ \ F[z_]\  = \ 
        F[z]\  + \ 
          R[[i
              ]]\/\((
                z - Z[[i]])\)\^p; ]; 

    For[i = 1, i ≤ Length[F[z]], \(i++\), 
\ \ \ 
      soset\  = \ 
        Solve[Denominator[\(F[z]\)[[i]]]
               == 0, z]; 
\ \ \ 
      Z[[i]]\  = \ 
        ReplaceAll[z, soset[[1]]]; 

      \ \ \ R[[i]]\  = \ 
        Residue[f[z], {z, Z[[i]]}]; ]; 

    Print["f[z] = ", f[z]]; 
Print["f[z] = ", F[z]]; \)

Determine which singularities lie inside C:  |z - i| = 1.

\(Print[\*""\!\(z\_1\) = "", 
      Z[[1]], ",    Is |", 
      Z[[1]], 
      " - I| < 1 ?    ", 
      N[Abs[Z[[1]] - \ I]] < 
        1]; 
Print[\*""\!\(z\_2\) = "", 
      Z[[2]], ",  Is |", 
      Z[[2]], 
      " - I| < 1 ?   ", 
      N[Abs[Z[[2]] - \ I]] < 
        1]; \)

Since only the first singularity lies inside C, add the first term in the sum forming  F[z]
times constant in the numerator that term times  2 π i.

\(val\  = \ 
      2  π\ I\ 
        Numerator[\(F[z]\)[[1]]]; 

    Print[\*""\!\(\(∫\_C\%.\)\)\!\(\(2  z\)\/\(z\^2\\  + \\ 2\)\)\[DifferentialD]z = "", val]; \)

Example 6.17,  Page 187.  Show that   (∫_C^( )) ((z - 2)/((z^2)  -  2))dz  =  - 6 π i ,
where  C  is the figure eight shown in the text.

\(Clear[f, F, i, p, R, solset, z, Z]; 

    f[z_]\  = \ \(z\  - \ 2\)\/\(z\^2\  - z\); 
Print["f[z] = ", f[z]]; 
    \)

First, split the function up into partial fractions involving linear terms in the denominators.  Don't worry about the subroutine. It splits f(z) up into partial fractions.  The result will be an equivalent function named  F[z].  The list of terms added to form F[z] have singularities in the same order in the list of points  Z.

\(solset\  = \ Sort[Solve[Denominator[f[z]] == 0, z]]; 

    Z\  = \ Table[0, {Length[solset]}]; 
R\  = \ Table[0, {Length[solset]}]; 
    
Clear[F]; \ F[z_]\  = \ 0; 

    For[i = 1, i ≤ Length[solset], \(i++\), 
\ \ \ 
      Z[[i]]\  = \ 
        ReplaceAll[z, solset[[i]]]; \ 

      \ \ \ If[1 == i, p\  = \ 1]; 
\ \ \ 
      If[1 < i; 
        Z[[i - 1]] == 
          Z[[i]], p\  = \ p + 1, 
        p\  = \ 1]; 
\ \ \ 
      R[[i]]\  = \ 
        Residue[\((z - Z[[i]])\)\^\(p - 1
                \)\ f[z], {z, Z[[i]]}]; 
      
\ \ \ F[z_]\  = \ 
        F[z]\  + \ 
          R[[i
              ]]\/\((
                z - Z[[i]])\)\^p; ]; 

    For[i = 1, i ≤ Length[F[z]], \(i++\), 
\ \ \ 
      soset\  = \ 
        Solve[Denominator[\(F[z]\)[[i]]]
               == 0, z]; 
\ \ \ 
      Z[[i]]\  = \ 
        ReplaceAll[z, soset[[1]]]; 

      \ \ \ R[[i]]\  = \ 
        Residue[f[z], {z, Z[[i]]}]; ]; 

    Print["f[z] = ", f[z]]; 
Print["f[z] = ", F[z]]; \)

Determine which singularities corresponds to the contour  (Subscript[C, 1])  and  (Subscript[C, 2]).  The first term will correspond to the singularity at  z = 1  used with contour (Subscript[C, 2]) and the second term corresponds to the singularity at  z = 0  used with contour (Subscript[C, 1]).

Since  (Subscript[C, 2])  has positive orientation, we add the first term in the sum forming  F(z)  times constant in the numerator of that term times  2π i.  And since the  (Subscript[C, 1])  has negative orientation, we subtract the second term in the sum forming  F(z)  times constant in the numerator of that term times  2π i.

\(val\  = \ 
      2\ π\ I\ 
          Numerator[\(F[z]\)[[1]]]\ \  - 
        \ 2\ π\ I\ 
          Numerator[\(F[z]\)[[2]]]; 

    Print[\*""\!\(\(∫\_C\%.\)\)\!\(\(z\\  - \\ 2\)\/\(z\^2\\  - z\)\)\[DifferentialD]z = "", val]; \)

GoTo Chapter 6

Section 6.4  The Fundamental Theorems of Integration

Let  f(z)  be analytic in the simply connected domain  D.  The theorems in this section show that an antiderivative  F(z)  can be constructed by contour integration.

Theorem 6.6,  Page 189.  (Indefinite Integrals or Antiderivatives)  Let  f(z)  be analytic in the simply connected domain  D.  If  (Subscript[z, 0])  is a fixed value in  D  and if  C  is any contour in  D  with initial point  (Subscript[z, 0])  and terminal point  z, then the function given by  F(z)=(∫_C^( )) f(z) dz=(∫_((Subscript[z, 0]))^z)f(ζ) dζ  is analytic in  D  and  (F )'(z)=f(z).

Theorem 6.7,  Page 192.  (Definite Integrals)  Let  f(z)  be analytic in a simply connected domain  D.  If  (Subscript[z, 0])  and  (Subscript[z, 1])  are two points in  D,  then   (∫_((Subscript[z, 0]))^((Subscript[z, 1])))f(z) dz=F((Subscript[z, 1]))-F((Subscript[z, 0])),   where  F(z)  is any antiderivative of  f(z).

Example 6.18,  Page 192.  Evaluate the definite integral.
(∫_( 1)^( i)) cos(z) dz   =  - sin(1)  + i sinh(1).

\(Clear[f, val1, val2, z]; 
f[z_]\  = \ Cos[z]; 

    g[z_]\  = ∫f[z] \[DifferentialD]z; 

    val1\  = \ g[I]\  - \ g[1]; 

    val2\  = \ ∫\_1\%I f[z] \[DifferentialD]z; 

    Print["    f[z] = ", f[z]]; 

    Print["    g[z] = ∫f[z]\[DifferentialD]z = ", g[z]]; 

    Print["    g[I] = ", g[I], ",   g[1] = ", 
      g[1]]; 
Print[
      \*""(∫_1^I)f[z]\[DifferentialD]z = g[I] - g[1] = "", val1]; 

    Print[\*""(∫_1^I)f[z]\[DifferentialD]z = "", val2]; 

    Print[\*""(∫_1^I)f[z]\[DifferentialD]z = "", N[val2]]; \)

Example 6.19,  Page 192.  Evaluate the definite integral.
(∫_( 0)^( 1)) 
            (e^(2 t)) 
              (e^(i π t/4)) 
              dt.

Clear[F, t];
F[t_]=(2 + ((I π)/4))∫((E^(2 t + I π t/4)))\[DifferentialD]t;
Print["(2+\!\(\(I\\ π\)\/4\))∫\!\(\*
StyleBox[\((\),
FontSize->15]\)\!\(\*
StyleBox[\(E\^\(2 t + I\\ π\\ t/4\)\),
FontSize->15]\))\[DifferentialD]t = ",F[t]];
Print["F[1]-F[0] = ", F[1] - F[0] ];
Print["F[1]-F[0] = ", Simplify[F[1] - F[0]]];
Print["(2+\!\(\(I\\ π\)\/4\))(∫_0^1)\!\(\*
StyleBox[\((\),
FontSize->15]\)\!\(\*
StyleBox[\(E\^\(2 t\)\),
FontSize->15]\)\!\(\*
StyleBox[\(E\^\(I\\ π\\ t/4\)\),
FontSize->15]\))\[DifferentialD]t = ",Simplify[F[1] - F[0]]];

Example 6.20,  Page 193.  Evaluate the definite integral.
(∫_( 4)^( 8 + 6 i)) (( (z^(-1/2)))/2)dz  =  1 + i.

\(Clear[f, val1, val2, z]; 
f[z_]\  = \ z\^\(\(-1\)/2\)\/2; 

    g[z_]\  = ∫f[z] \[DifferentialD]z; 

    val1\  = \ g[8 + 6  I]\  - \ g[4]; 

    val2\  = \ 
      ∫\_4\%\(8 + 6  I\)f[z] \[DifferentialD]z; 

    Print["       f[z] = ", f[z]]; 

    Print["       g[z] = ∫f[z]\[DifferentialD]z = ", g[z]]; 

    Print["    g[8+6I] = ", g[8 + 6  I], 
      ",   g[4] = ", g[4]]; 

    Print[\*""(∫_4^(8 + 6  I))f[z]\[DifferentialD]z = g[8+6I] - g[4] = "", val1]; 

    Print[\*""(∫_4^(8 + 6  I))f[z]\[DifferentialD]z = "", val2]; 

    Print[\*""(∫_4^(8 + 6  I))f[z]\[DifferentialD]z = "", N[val2]]; \)

Example 6.21 and 6.22,  Page 193.  Evaluate the definite integral  
(∫_C^( )) (1/z)dz = 2 π i , where  C  is the unit circle  C: |z| = 1, taken with positive orientation.

\(Clear[f, g, G, lim2, lim1, θ, z, Z]; 
f[Z_]\  = \ 1\/Z; 

    g[z_]\  = \ ∫f[z] \[DifferentialD]z; 

    z[θ_]\  = \ E\^\(I\ θ\); 

    lim2\  = \ 
      Limit[g[z[θ]], θ -> π, Direction -> 1]; 

    lim1\  = \ 
      Limit[g[z[θ]], θ -> \(-π\), 
        Direction -> \(-1\)]; 
Print["f[z] = ", f[z]]; 

    Print["g[z] = ∫f[z]\[DifferentialD]z  = ", g[z]]; 

    Print[\*""lim2 = \!\(lim\+\(θ -> \\ π\)\) Log[\!\(E\^\(I\\ θ\)\)] = "", lim2]; 

    Print[\*""lim1 = (lim_(θ -> -π)) Log[\!\(E\^\(I\\ θ\)\)] = "", lim1]; 

    Print[\*""\!\(\(∫\_C\%.\)\)(1/z)\[DifferentialD]z = lim2 - lim1 = "", lim2 - lim1]; \)

Another approach is to parameterize the integrand.

\(w\  = \ f[z[θ]] \(z'\)[θ]; 

    G[θ_]\ \  = \ ∫w\ \[DifferentialD]θ; 

    val\  = \ G[π]\  - \ G[\(-π\)]; 

    Print["         z[θ] = ", \ z[θ]]; 

    Print["        z'[θ] = ", \ \(z'\)[θ]]; 

    Print["         f[z] = ", \ f[z]]\ ; 

    Print[" f[z[θ]]z'[θ] = ", \ w]; 

    Print["         G[θ] = ∫f[z[θ]]z'[θ] = ", \ G[θ]]; 

    Print["         G[π] = ", \ G[π], \ ",  G[-π] = ", 
      G[\(-π\)]]; 

    Print[\*""       \!\(\(∫\_C\%.\)\)(1/z)\[DifferentialD]z = G[π] - G[-π] = "", val]; \)

GoTo Chapter 6

Section 6.5  Integral Representations for Analytic Functions

We now present some major results in the theory of functions of a complex variable.
The first result is known as Cauchy's integral formula and shows that the value of an
analytic function  f  can be represented by a certain contour integral.

Theorem 6.8,  Page 195. (Cauchy's Integral Formula)  Let  f(z)  be analytic in the simply connected domain  D,  and let  C  be a simple closed positively oriented contour that lies in  D.  If  (Subscript[z, 0]) is a point that lies interior to  C,  then  f((Subscript[z, 0]))=(1/(2 π i))(∫_C^( )) ((f(z))/(z  -  (Subscript[z, 0])))dz.

Theorem 6.9,  Page 197. (Leibniz's Rule)  Let  D  be a simply connected domain,  and let  I :  a ≤ t ≤ b  be an interval of real numbers.  Let  f(z,t)  and its partial derivative  (∂_z) f(z, t)  with respect to  z  be continuous functions for all  z  in  D  and all  t  in  I.  Then 
F(z)= (∫_a^b)f(z, t) dt   is analytic for  z  in  D,  and   (F^( ′))(z)= (∫_a^b)(∂_z) f(z, t) dt.

Theorem 6.10,  Page 197. (Cauchy's Integral Formulae for Derivatives)  Let  f(z)  be analytic in the simply connected domain  D,  and let  C  be a simple closed positively oriented contour that lies in  D.  If  z  is a point that lies interior to  C,  then   (f^((n)))(z)=((n!)/(2 π i))(∫_C^( )) ((f(ζ))/((((z - ζ))^(n + 1))))dζ.

Corollary 6.1,  Page 199.  If  f(z)  is analytic in the domain  D,  then all derivatives (f^( ′))(z),(f^( ′))'(z),⋯,(f^((n)))(z),⋯  exists and are analytic in  D.

Corollary 6.2,  Page 199.  If  u(x, y)  is a harmonic function at each point  (x,y)  in the domain  D,  then all partial derivatives  (∂_x) u(x, y), 
      (∂_y) u(x, y), (∂_(x, x))u(x, y), 
      (∂_(x, y))u(x, y), (∂_(y, y))u(x, y)  exists and are harmonic functions.

Example 6.23,  Page 196.  Show that   (∫_C^( )) ((exp(z))/(z  -  1))dz  =  i 2 π e,
where  C  is the circle C: |z| = 2  with positive orientation.

\(Clear[f, i, solset, val, w, z, Z]; 

    w\  = \ E\^z\/\(z - 1\); 

    solset\  = \ Sort[Solve[Denominator[w] == 0, z]]; 

    Z\  = \ Table[0, {Length[solset]}]; 

    For[i = 1, i ≤ Length[solset], \(i++\), 
      \(Z[[i]]\  = \ 
        ReplaceAll[z, solset[[i]]]; \)]; 
    
Print["w = ", w]; 
Print["The set of singularities is ", Z]; 
    \)

Find the singularity that lie inside C.

\(\(Print[\*""\!\(z\_1\) = "", 
      Z[[1]], 
",    Is |", 
      Z[[1]], "| < 2 ?    ", 
      N[Abs[Z[[1]]]] < 2]; \)\)

The integrand f(z) is to be used is:

\(f[z_]\  = \ 
      w\ \((z\  - \ Z[[1]])\); 

    Print["f[z] = ", f[z]]; \)

The integral of f(z) taken over C is:

\(val\  = \ 
      2\ π\ I\ 
        f[Z[[1]]]; 

    Print[" f[z] = ", f[z]]; 

    Print[\*""   (Subscript[z, 1]) = "", 
      Z[[1]]]; 

    Print[\*""f[(Subscript[z, 1])] = "", \ 
      f[Z[[1]]]]; 

    Print[\*""\!\(\(∫\_C\%.\)\)((E^z)/(z - 1))\[DifferentialD]z = 2πI f[(Subscript[z, 1])] = "", val]; \)

Example 6.24,  Page 196.  Show that   (∫_C^( )) ((sin(z))/(4 z  +  π))dz  =  ((- (Sqrt[2]) π i)/4) ,
where  C  is the circle C: |z| = 1  with positive orientation.

\(Clear[f, i, solset, val, w, z, Z]; 

    w\  = \ Sin[z]\/\(4  z\  + \ π\); 

    solset\  = \ Sort[Solve[Denominator[w] == 0, z]]; 

    Z\  = \ Table[0, {Length[solset]}]; 

    For[i = 1, i ≤ Length[solset], \(i++\), 
      \(Z[[i]]\  = \ 
        ReplaceAll[z, solset[[i]]]; \)]; 
    
Print["w = ", w]; 
Print["The set of singularities is ", Z]; 
    \)

Find the singularity that lie inside C.

\(\(Print[\*""\!\(z\_1\) = "", 
      Z[[1]], 
",    Is |", 
      Z[[1]], "| < 1 ?    ", 
      N[Abs[Z[[1]]]] < 1]; \)\)

The integrand f(z) is to be used is:

\(f[z_]\  = \ 
      Simplify[w\ \((z\  - \ Z[[1]])\)]; 
    
Print["f[z] = ", f[z]]; \)

The integral of f(z) taken over C is:

\(val\  = \ 
      2\ π\ I\ 
        f[Z[[1]]]; 

    Print[" f[z] = ", f[z]]; 

    Print[\*""   (Subscript[z, 1]) = "", 
      Z[[1]]]; 

    Print[\*""f[(Subscript[z, 1])] = "", \ 
      f[Z[[1]]]]; 

    Print[\*""\!\(\(∫\_C\%.\)\) \!\(Sin[z]\/\(4  z\\  + \\ π\)\)\[DifferentialD]z =  2πI f[(Subscript[z, 1])] = "", 
      val]; \)

Example 6.25,  Page 198.   Let  (Subscript[z, 0])  denote a fixed complex value.   If  C  is a 
simple closed positively orientated contour such that  (Subscript[z, 0])  lies interior to  C, then

(∫_C^( )) (1/(z - (Subscript[z, 0])))dz  =  2 π i ,

(∫_C^( )) (1/(((z - (Subscript[z, 0])))^n))dz  =  0 , where  n ≠ 1 is an integer.

(a)   Show that   (∫_C^( )) (1/(z - (Subscript[z, 0])))dz  =  2 π i .
For illustration, we use a circle of radius 1 centered at  (Subscript[z, 0]).

\(Clear[f, F, θ, z, z0]; 
f[z_]\  = \ 1\/\(z - z0\); 

    z[θ_]\  = \ z0\  + \ E\^\(I\ θ\); 
    
F[θ_]\  = 
      ∫f[z[θ]] \(z'\)[θ] \[DifferentialD]θ; 

    val\  = \ ∫\_0\%\(2  π\)f[z[θ]] \(z'\)[θ] 
          \[DifferentialD]θ; 

    Print["              f[z] = ", f[z]]; 

    Print["              z[θ] = ", z[θ]]; 

    Print["           f[z[θ]] = ", f[z[θ]]]; 

    Print["             z'[θ] = ", \(z'\)[θ]]; 

    Print["     f[z[θ]] z'[θ] = ", 
      f[z[θ]]\ \(z'\)[θ]]; 

    Print["   ∫f[z[θ]]z'[θ]\[DifferentialD]t = ", 
      F[θ]]; 

    Print["F[2π] = ", F[2  π], ",    F[0] = ", F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = F[2π]-F[0] = "", 
F[2  π] - F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = "", val]; \)

(b)   Show that   (∫_C^( )) (1/(((z - (Subscript[z, 0])))^5))dz  =  0 ,
For illustration, we use a circle of radius 1 centered at  (Subscript[z, 0]).

Clear[f, F, θ, z, z0]\(f[z_]\  = \ 1\/\((z - z0)\)\^5; 

    z[θ_]\  = \ z0\  + \ E\^\(I\ θ\); 
    
F[θ_]\  = 
      ∫f[z[θ]] \(z'\)[θ] \[DifferentialD]θ; 

    val\  = \ ∫\_0\%\(2  π\)f[z[θ]] \(z'\)[θ] 
          \[DifferentialD]θ; 

    Print["              f[z] = ", f[z]]; 

    Print["              z[θ] = ", z[θ]]; 

    Print["           f[z[θ]] = ", f[z[θ]]]; 

    Print["             z'[θ] = ", \(z'\)[θ]]; 

    Print["     f[z[θ]] z'[θ] = ", 
      f[z[θ]]\ \(z'\)[θ]]; 

    Print["   ∫f[z[θ]]z'[θ]\[DifferentialD]t = ", 
      F[θ]]; 

    Print["F[2π] = ", F[2  π], ",    F[0] = ", F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = F[2π]-F[0] = "", 
F[2  π] - F[0]]; 

    Print[\*"" (∫_0^(2  π))f[z[θ]]z'[θ]\[DifferentialD]t = "", val]; \)

Example 6.26,  Page 198.  Show that   (∫_C^( )) ((exp((z^2)))/((( z  -  i ))^4))dz  =  ((- 4 π)/(3 e)),
where  C  is the circle C: |z| = 2  with positive orientation.

\(Clear[f, i, solset, val, w, z, Z]; 

    w\  = \ E\^\(z\^2\)\/\((z\  - \ I)\)\^4; 

    solset\  = \ Sort[Solve[Denominator[w] == 0, z]]; 

    Z\  = \ Table[0, {Length[solset]}]; 

    For[i = 1, i ≤ Length[solset], \(i++\), 
      \(Z[[i]]\  = \ 
        ReplaceAll[z, solset[[i]]]; \)]; 
    
Print["w = ", w]; 
Print["The set of singularities is ", Z]; 
    \)

Find the singularity that lie inside C.

\(\(Print[\*""\!\(z\_1\) = "", 
      Z[[1]], 
",    Is |", 
      Z[[1]], "| < 2 ?    ", 
      N[Abs[Z[[1]]]] < 2]; \)\)

Since  z = i  is a singularity of order  n = 4 , multiply the integrand by  (((z  -  i))^4)  to obtain the function  f(z).

\(f[z_]\  = \ 
      w\ \((z\  - \ Z[[1]])\)\^4; 

    Print["f[z] = ", f[z]]; \)

The integral of f(z) taken over C is obtained by applying the Cauchy integral formula for derivatives.

\(f3\  = \ Simplify[Expand[\(\(\(f'\)'\)'\)[z]]]; 

    val\  = \ \(2\ π\ I\)\/\(3!\)\ 
        \(\(\(f'\)'\)'\)[Z[[1]]]; 

    Print[" f'''[z] = ", f3]; 

    Print[\*""      (Subscript[z, 1]) = "", 
      Z[[1]]]; 

    Print[\*""f'''[(Subscript[z, 1])] = "", \ 
      \(\(\(f'\)'\)'\)[Z[[1]]]]; 

    Print[\*""\!\(\(∫\_C\%.\)\) \!\(E\^\(z\^2\)\/\((z\\  - \\ I)\)\^4\)\[DifferentialD]z = \!\(\(2\\ π\\ I\)\/\(3!\)\) f'''[(Subscript[z, 1])] = "", val]; \)

Example 6.27,  Page 198.  Show that   (∫_C^( )) ((exp( i π z))/(2 (z^2)  -  5 z  +  2))dz  =  ((2 π)/3),
where  C  is the circle C: |z| = 1  with positive orientation.

\(Clear[f, i, solset, val, w, z, Z]; 

    w\  = \ E\^\(I\ π\ z\)\/\(2  z\^2\  - \ 
          5  z\  + \ 2\); 
solset\  = \ Sort[Solve[Denominator[w] == 0, z]]; 
    
Z\  = \ Table[0, {Length[solset]}]; 

    For[i = 1, i ≤ Length[solset], \(i++\), 
      \(Z[[i]]\  = \ 
        ReplaceAll[z, solset[[i]]]; \)]; 
    
Print["w = ", w]; 
Print["The set of singularities is ", Z]; 
    \)

Find the singularity that lie inside C.

\(\(Print[\*""\!\(z\_1\) = "", 
      Z[[1]], 
",    Is |", 
      Z[[1]], "| < 1 ?    ", 
      N[Abs[Z[[1]]]] < 1]; 

    Print[\*""\!\(z\_2\) = "", 
      Z[[2]], 
",    Is |", 
      Z[[2]], "| < 1 ?    ", 
      N[Abs[Z[[2]]]] < 1]; 
\)\)

Thus,  z = (1/2)  the only singularity that lies inside C.

The integrand f(z) is to be used is:

\(f[z_]\  = \ 
      Simplify[w\ \((z\  - \ Z[[1]])\)]; 
    
Print["f[z] = ", f[z]]\)

\(val\  = \ 
      2\ π\ I\ 
        f[Z[[1]]]; 

    Print[" f[z] = ", f[z]]; 

    Print[\*""   (Subscript[z, 1]) = "", 
      Z[[1]]]; 

    Print[\*""f[(Subscript[z, 1])] = "", \ 
      f[Z[[1]]]]; 

    Print[\*""\!\(\(∫\_C\%.\)\) \!\(E\^\(I\\ π\\ z\)\/\(2  z\^2\\  - \\ 5  z\\  + \\ 2\)\)\[DifferentialD]z =  2πI f[(Subscript[z, 1])] = "", val]; \)

GoTo Chapter 6

Section 6.6  The Theorems of Morera and Liouville and Some Applications

Theorem 6.11,  Page 201. (Morera's Theorem)  Let  f  be a continuous function in a simply connected domain D.  If  (∫_C^( )) f(z) dz=0  for every closed contour in  D,  then  f  is analytic in  D.

Theorem 6.12,  Page 201. (Gauss's Mean Value Theorem)  If  f  is analytic in a simply connected domain  D  that contains the circle  \(TraditionalForm
      \`\(C : \)\  | \(z - z\_0 | \) = R
        \),  then  f((Subscript[z, 0]))=(1/(2 π)) (∫_0^(2  π))f((Subscript[z, 0]) + R (e^i)θ) 
                dθ.

Theorem 6.13,  Page 202. (Maximum Modulus Principle)  Let  f  be analytic and nonconstant in the domain  D.  Then  \(TraditionalForm
      \`\(|\( f(z) | \)\)\)  does not attain a maximum value at any point  (Subscript[z, 0])  in  D.

Theorem 6.14,  Page 204. (Cauchy's Inequalities)  Let  f  be analytic and nonconstant in the domain  D  that contains the circle  \(TraditionalForm
      \`\(C : \)\  | \(z - z\_0 | \) = R
        \).  If  \(TraditionalForm
      \`| f(z) | \( < M\)\)  holds for all points  z  on  C,  then  ∣(f^((n)))((Subscript[z, 0]))∣≤((n! M)/((R^n)))   for   n = 1, 2, ⋯ .

Theorem 6.15,  Page 205. (Liouville's Theorem)  If  f  is an entire function and is bounded for all values of  z  in the complex plane,  then  f  is constant.

Theorem 6.16,  Page 205. (The Fundamental Theorem of Algebra)  If  P(z)  is a polynomial of degree  n,  then  P  has at least one zero.

Corollary 6.3,  Page 206.  Let  P  be a polynomial of degree  n.  Then  P  can be expressed as the product of linear factors.  That is,  \(TraditionalForm
      \`P(z) = A\ \((z - z\_1)\) \((z - z\_2)\) 
          \(⋯(z - z\_n)\)\)  where  (Subscript[z, 1]), (Subscript[z, 2]), ⋯, (Subscript[z, n])  
are the zeros of  P  counted according to multiplicity an  A  is a constant.

Example 6.28,  Page 204.  Let  f(z) = az + b,  where the domain is the disk  \(TraditionalForm
      \`\(D : \)\  | 
        \(z | \( < \ 1\)\)\).
Then  f(z)  is continuous on the closed region  \(TraditionalForm
      \`\(R : \)\  | 
        \(z | \( ≤ \ 1\)\)\)  and   (max_( ∣ z ∣  < 1))| f(z) | =  | a | 
            +| b | ,  and this value is assumed by  f(z)  at a point  (Subscript[z, 0])=(e^(i (Subscript[θ, 0])))  on the boundary of  D.
Solution.  The solution is given in the textbook.

Example 6.29,  Page 205.  Show that the function   sin z   is NOT a bounded function.

Clear[f, F, y, z]f[z_]  =  Sin[z];
F1[y_]  =  Sin[(π/2)  +  I y];
F2[y_]  =  TrigToExp[F1[y]];
lim1  =  Limit[ F1[y],  y -> ∞];
lim2  =  Limit[ F2[y],  y -> ∞];
Print["    f[z] = ", f[z]];
\(Print[\*""f[\!\(π\/2\)+Iy] = "", 
        f[π\/2\  + \ I\ y]]\);
\(Print[\*""    \!\(F\_1\)[y] = "", F1[y]]\);
\(Print[\*""    \!\(F\_2\)[y] = "", F2[y]]\);
Print["\!\(lim\+\(y -> 
\*InterpretationBox[\(∞\),
DirectedInfinity[ 1]]\)\) (Subscript[F, 1])[y] = ",lim1];
Print["\!\(lim\+\(y -> 
\*InterpretationBox[\(∞\),
DirectedInfinity[ 1]]\)\) (Subscript[F, 2])[y] = ",lim2];

GoTo Chapter 6