COMPLEX ANALYSIS: Mathematica 4.1 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 2002
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
4th Edition,  2001,  ISBN: 0-7637-1425-9
Jones & Bartlett Publishers, Inc.  
40 Tall Pine Drive, Sudbury, MA 01776  
Tele.  (800) 832-0034, FAX:  (508)  443-8000  
E-mail:  mkt@jbpub.com  
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This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 6  COMPLEX INTEGRATION

a = b = c = d = R = t = u = U = v = V = w = W = x = y = z = Z = 
                                      0; 
Clear[a, b, c, center, centers, d, dot, dots, f, F, i, int, int1, 
        int2, Int, ImInt, ImVal, label, path, r, R, ReInt, ReVal, soset, 
        solset, t, u, U, U1, U2, v, V, V1, V2, val, Val, w, W, W1, W2, Wint, 
        x, x0, x1, y, y0, y1, z, z1, z2, Z, Z1, Z2, 
        Z3]; 
Needs["Graphics`Arrow`"]; 
Needs["Graphics`Colors`"];

Section 6.1	Complex Integrals

Section 6.2	Contours and Contour Integrals

Section 6.3	The Cauchy-Goursat Theorem

Section 6.4	The Fundamental Theorem of Integration

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.  The  (n^th)  derivative,  (f^((n))) (z),  will have a similar representation.  In Chapter 7 we will show how the Cauchy integral formulae are used to prove Taylor's theorem, and we will establish the power series representation for analytic functions.  The Cauchy integral formulae will also be a convenient tool for evaluating certain integrals.

Theorem 6.8 (Cauchy Integral Formula), Page 244.  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 \((z\_0)\) = \(1\/\(2  π\ i\)\) \(∫\_C\(\( f \((z)\)\)\/\(z\  - \ z\_0\)\) dz\)\).

Proof of Theorem 6.8, see text Page 244.

Example 6.21,  Page 245.  Show that  (∫_C) ((e^z)/(z  -  
                  1)) dz = 
        2 i π e,  where C is the circle  C : |z| = 2  with positive orientation.

Solution 6.21.

Enter the integrand  w[z] = ((e^z)/(z  -  1))  and locate the singularities.

Remove[w, z, Z]; 
Clear[f, i, solset, val]; 
w[z_]  =  ((E^z)/(z - 1)); 
solset  =  
        Sort[Solve[Denominator[w[z]] == 0, z]]; 
Z  =  Table[0, {Length[solset]}]; 
For[i = 1, 
        i ≤ 
          Length[solset], i++, 
(Subscript[Z, [[i]]])  =  
            ReplaceAll[z, 
              (Subscript[solset, [[i]]])]; ]; 
Print["w[z] = ", w[z]]; 
Print["The set of singularities is ", 
        Z];

Find the singularity that lie inside  C : |z| = 2.

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

The function f(z) in the numerator

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

Use Cauchy's Integral Formula to evaluate the integral of  w[z] = ((e^z)/(z  -  1)) taken over C.

val  =  
        2 π I f[
            (Subscript[Z, [[1]]])]; 
\(\(Print["w[z] = ", "f[z]"\/\(z\  - \ Z\_\(\([[\)\(1\)\(]]\)\)\)];\)\ \)
Print[""]; 
Print[" f[z] = ", f[z]]; 
\(\(Print[\*""   (Subscript[z, 1]) = "", 
        Z\_\(\([[\)\(1\)\(]]\)\)];\)\ \)
\(\(Print[\*""f[(Subscript[z, 1])] = "", \ 
        f[Z\_\(\([[\)\(1\)\(]]\)\)]];\)\ \)
Print[""]; 
Print["Use Cauchy's Integral Formula"]; 
\(\(Print[\*""(∫_C)((f (z))/(z - (Subscript[z, 1])))\[DifferentialD]z = 2πI f((Subscript[z, 1]))""];\)\ \)
\(\(Print[\*""(∫_C) "", 
        w[z], " \[DifferentialD]z = 2πI f[", 
        Z\_\(\([[\)\(1\)\(]]\)\), "] = 2πI(", 
        f[Z\_\(\([[\)\(1\)\(]]\)\)], ") = ", val];\)\ \)

Thus, we have found the value of the contour integral.

dots = 
        Graphics[{Red, PointSize[0.03], 
            Point[{Re[
                  (Subscript[Z, [[1]]])], Im[(Subscript[Z, [[1]]])]}]}]; 
\(\(centers = 
        Graphics[{Text[\*""(Subscript[z, 1])"", {Re[
                  Z\_\(\([[\)\(1\)\(]]\)\)], Im[Z\_\(\([[\)\(1\)\(]]\)\)] + 
                  0.2}, {\(-1\), 0}]}];\)\ \)
contour  =  Graphics[{Magenta, Circle[{0, 0}, 2]}]; 
Show[dots, centers, contour, 
        PlotRange -> {{-2.1, 2.1}, {-2.1, 2.1}}, 
        Ticks -> {Range[-2, 2, 1], 
            Range[-2, 2, 1]}, 
AspectRatio -> 1, 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(Print[\*""\!\(∮\+C\) ("", 
      w[z], ") \[DifferentialD]z = ", val]\)
Print["The contour is the circle  C: |z| = 2,"]; 
Print["with positive orientation."];

Example 6.22,  Page 245.  Show that  \(∫\_C\(\( 
                  sin \((z)\)\)\/\(4  
                    z\  + \ \ π\)\) dz = \(\(-i\)\ π \@ 2\)\/4\),  where C is the circle  C : |z| = 1  with positive orientation.

Solution 6.22.

Enter the integrand  w[z] = Sin[z]/(4  z  +  π)  and locate the singularities.

Remove[w, z, Z]; 
Clear[f, i, solset, val]; 
w[z_]  =  Sin[z]/(4  z  +  π); 
solset  =  
        Sort[Solve[Denominator[w[z]] == 0, z]]; 
Z  =  Table[0, {Length[solset]}]; 
For[i = 1, 
        i ≤ 
          Length[solset], i++, 
(Subscript[Z, [[i]]])  =  
            ReplaceAll[z, 
              (Subscript[solset, [[i]]])]; ]; 
Print["w[z] = ", w[z]]; 
Print["The set of singularities is ", 
        Z];

Find the singularity that lie inside  C : |z| = 1.

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

Use Cauchy's Integral Formula to evaluate the integral of  w[z] = Sin[z]/(4  z  +  π) taken over C.

val  =  
        2 π I f[
            (Subscript[Z, [[1]]])]; 
\(\(Print["w[z] = ", "f[z]"\/\(z\  - \ Z\_\(\([[\)\(1\)\(]]\)\)\)];\)\ \)
Print[""]; 
Print[" f[z] = ", f[z]]; 
\(\(Print[\*""   (Subscript[z, 1]) = "", 
        Z\_\(\([[\)\(1\)\(]]\)\)];\)\ \)
\(\(Print[\*""f[(Subscript[z, 1])] = "", \ 
        f[Z\_\(\([[\)\(1\)\(]]\)\)]];\)\ \)
Print[""]; 
Print["Use Cauchy's Integral Formula"]; 
\(\(Print[\*""(∫_C)((f (z))/(z - (Subscript[z, 1])))\[DifferentialD]z = 2πI f((Subscript[z, 1]))""];\)\ \)
\(\(Print[\*""(∫_C) "", 
        w[z], " \[DifferentialD]z = 2πI f[", 
        Z\_\(\([[\)\(1\)\(]]\)\), "] = 2πI(", 
        f[Z\_\(\([[\)\(1\)\(]]\)\)], ") = ", val];\)\ \)

Thus, we have found the value of the contour integral.

dots = 
        Graphics[{Cyan, PointSize[0.03], 
            Point[{Re[
                  (Subscript[Z, [[1]]])], Im[(Subscript[Z, [[1]]])]}]}]; 
\(\(centers = 
        Graphics[{Text[\*""(Subscript[z, 1])"", {Re[
                  Z\_\(\([[\)\(1\)\(]]\)\)], Im[Z\_\(\([[\)\(1\)\(]]\)\)] + 
                  0.1}, {\(-1\), 0}]}];\)\ \)
contour  =  Graphics[{Blue, Circle[{0, 0}, 1]}]; 
Show[dots, centers, contour, 
        PlotRange -> {{-1.1, 1.1}, {-1.1, 1.1}}, 
        Ticks -> {Range[-1, 1, 1], 
            Range[-1, 1, 1]}, 
AspectRatio -> 1, 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(Print[\*""\!\(∮\+C\) ("", 
      w[z], ") \[DifferentialD]z = ", val]\)
Print["The contour is the circle  C: |z| = 1,"]; 
Print["with positive orientation."];

Example 6.23,  Page 245.  Show  (∫_C) ((e^(i π z))/(2 
                     (z^2)  -  5  z  +  
                  2)) dz = ((2 π)/3),  where C is the circle  C : |z| = 1  with positive orientation.

Solution 6.23.

Enter the integrand  \(w[
          z] = e\^\(i\ π\ z\)\/\(2  z\^2\  - \ 5  z\  + \ 2\)\)  and locate the singularities.

Remove[w, z, Z]; 
Clear[f, i, solset, val]; 
w[
          z_]  =  ((E^(I π z))/(2  (z^2)  -  5  z  +  2)); 
solset  =  
        Sort[Solve[Denominator[w[z]] == 0, z]]; 
Z  =  Table[0, {Length[solset]}]; 
For[i = 1, 
        i ≤ 
          Length[solset], i++, 
(Subscript[Z, [[i]]])  =  
            ReplaceAll[z, 
              (Subscript[solset, [[i]]])]; ]; 
Print["w[z] = ", w[z]]; 
Print["The set of singularities is ", 
        Z];

Find the singularity that lie inside C.

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

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

The integrand f(z) is to be used is

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

Use Cauchy's Integral Formula to evaluate the integral of  \(w[
          z] = e\^\(i\ π\ z\)\/\(2  z\^2\  - \ 5  z\  + \ 2\)\) taken over C.

val  =  
        2 π I f[
            (Subscript[Z, [[1]]])]; 
\(\(Print["w[z] = ", "f[z]"\/\(z\  - \ Z\_\(\([[\)\(1\)\(]]\)\)\)];\)\ \)
Print[""]; 
Print[" f[z] = ", f[z]]; 
\(\(Print[\*""   (Subscript[z, 1]) = "", 
        Z\_\(\([[\)\(1\)\(]]\)\)];\)\ \)
\(\(Print[\*""f[(Subscript[z, 1])] = "", \ 
        f[Z\_\(\([[\)\(1\)\(]]\)\)]];\)\ \)
Print[""]; 
Print["Use Cauchy's Integral Formula"]; 
\(\(Print[\*""(∫_C)((f (z))/(z - (Subscript[z, 1])))\[DifferentialD]z = 2πI f((Subscript[z, 1]))""];\)\ \)
\(\(Print[\*""(∫_C) "", 
        w[z], \*"" \[DifferentialD]z = (∫_C) "", "(", f[z], ")/(", 
        z\  - \ Z\_\(\([[\)\(1\)\(]]\)\), ")", " \[DifferentialD]z"];\)\ \)
\(\(Print[\*""(∫_C) "", 
        w[z], " \[DifferentialD]z = 2πI f[", 
        Z\_\(\([[\)\(1\)\(]]\)\), "] = 2πI(", 
        f[Z\_\(\([[\)\(1\)\(]]\)\)], ") = ", val];\)\ \)

Thus, we have found the value of the contour integral.

dots = 
        Graphics[{Cyan, PointSize[0.03], 
            Point[{Re[
                  (Subscript[Z, [[1]]])], Im[(Subscript[Z, [[1]]])]}]}]; 
\(\(centers = 
        Graphics[{Text[\*""(Subscript[z, 1])"", {Re[
                  Z\_\(\([[\)\(1\)\(]]\)\)], Im[Z\_\(\([[\)\(1\)\(]]\)\)] + 
                  0.1}, {\(-1\), 0}]}];\)\ \)
contour  =  Graphics[{Blue, Circle[{0, 0}, 1]}]; 
Show[dots, centers, contour, 
        PlotRange -> {{-1.1, 1.1}, {-1.1, 1.1}}, 
        Ticks -> {Range[-1, 1, (1/2)], 
            Range[-1, 1, (1/2)]}, 
AspectRatio -> 
          1, Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(Print[\*""\!\(∮\+C\) ("", 
      w[z], ") \[DifferentialD]z = ", val]\)
Print["The contour is the circle  C: |z| = 1,"]; 
Print["with positive orientation."];

We now state a general result that shows how differentiation under the integral sign can be accomplished. The proof can be found in some advanced texts. See, for instance, Rolf Nevanlinna and V. Paatero, Introduction to Complex Analysis (Reading, Massachusetts: Addison-Wesley Publishing Company, 1969), Section 9.7.

Theorem 6.11 (Leibniz's Rule), Page 246.  Let  G  be an open set,  and let  I :  a ≤ t ≤ b  be an interval of real numbers.  Let g(z,t) and its partial derivative (Subscript[g, z]) (z, t) with respect to z be continuous functions for all z in G and all t in I.  Then  

			F (z) = (∫_a^b) g (z, t) dt  is analytic for z in G, and  

			F' (z) = (∫_a^b) (Subscript[g, z]) (z, 
              t) dt.

Proof of Theorem 6.11, see text Page 246.

We now generalize Theorem 6.10 to give an integral representation for the (n^th) derivative, (f^((n))) (z). We use Leibniz's rule in the proof, and we shall see that this method of proof will be a mnemonic device for remembering the upcoming theorem.

Demonstration for Theorem 6.11, see text Page 197.

Remove[t, z]; 
Clear[f, F]; 
F[
          z_] = (∫_a^b) f[z, 
              t] \[DifferentialD]t; 
Print["F[z] = ", F[z]]; 
Print[""]; 
Print["Rule Leibniz's Rule is"]; 
Print["F'[z] = ", F'[z]]; 
Print["where"]; 
\(\(Print[∂\_z\ 
          f[z, t], \*"" = \!\(∂\_z\) f[z,t]""];\)\ \)

Theorem 6.10 (Cauchy's Integral Formulae for Derivatives), Page 246.  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 for any integer n ≥ 0, we have

			\(\(f\^\((n)\)\) \((z)\) = \(\(n!\)\/\(2  π\ i\)\) \(∫\_C\(\( 
                    f \((ζ)\)\)\/\((ζ - z)\)\^\(n + 1\)\) dζ\)\).

Proof of Theorem 6.10, see text Page 256.

Example 6.24,  Page 247.  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  (a)  (∫_C) (1/(z  -  (Subscript[z, 0]))) dz = 
        2  π i,  and  (b)  (∫_C) (1/(((z  -  (Subscript[z, 0])))^m)) dz = 
        0   where  m ≠ 1  is an ingeger.

Solution 6.24 (a).

(a)   Show that (∫_C) (1/(z  -  (Subscript[z, 0]))) dz = 
        2  π i.  Use the Cauchy Integral Formula in the form (∫_C)(( f (z))/(z  -  
                  (Subscript[z, 0]))) dz = 
        2  π i f ((Subscript[z, 0])).

Remove[z, w]; 
Clear[f]; 
f[z_] = 1; 
z0 = (Subscript[z, 0]); 
w[z_]  =  (1/(z - z0)); 
Print["f[z] = ", f[z]]; 
\(\(Print[\*""f[\!\(z\_0\)] = "", 
        f[z0]];\)\ \)
Print[""]; 
Print["Use Cauchy's Integral Formula"]; 
\(\(Print[\*""(∫_C)((f (z))/(z - (Subscript[z, 0])))\[DifferentialD]z = 2πI f((Subscript[z, 0])) = 2πI("", f[z0], ")"];\)\ \)
\(\(Print[\*""(∫_C)((f (z))/(z - (Subscript[z, 0])))\[DifferentialD]z = "", 
        2\ π\ I\ f[z0]];\)\ \)

Thus, we have found the value of the contour integral.

dot = 
        Graphics[{Red, PointSize[0.03], 
            Point[{2, (5/3)}]}]; 
\(\(center = 
        Graphics[{Text[\*""\!\(z\_0\)"", {2, 1.6}, {0, 
                1}]}];\)\ \)
contour  =  
        Graphics[{Magenta, Circle[{2, (5/3)}, 1]}]; 
Show[center, contour, dot, PlotRange -> {{0, 4}, {0, 3}}, 
        Ticks -> None, 
AspectRatio -> (3/4), 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(Print[\*""\!\(∮\+C\) "", 
      w[z], " \[DifferentialD]z = ", 
      2\ π\ I\ f[z0]]\)
Print["For illustration,"]; 
\(\(Print[\*""a contour could be the unit circle C: |z - (Subscript[z, 0])| = r,""];\)\ \)
Print["with positive orientation."];

Solution 6.24 (b).

(b)   Show that  (∫_C) (1/(((z  -  (Subscript[z, 0])))^m)) dz = 
        0,  when  m ≠ 1  is an integer. 
Remark.  If  m  is an integer and  m < 0,  then  m = -n,  where  n  is a positive integer and the integrand is  f (z) = (((z - (Subscript[z, 0])))^n)
 which is a polynomial of degree n.  Hence in this case, f(z) would be analytic and the Cauchy-Goursat theorem implies that the value of the integral is zero.
 
 Now we consider the case when m is a positive integer and  m > 1.
 Use the Cauchy's Integral Formulae for Derivatives in the form  \(∫\_C\(\( 
                  f \((z)\)\)\/\((z - z\_0)\)\^\(n + 1\)\) dz = \(\(2  π\ i\)\/\(n!\)\) \(f\^\((n)\)\) \((z\_0)\)\)
 
 For illustration.  We use m = 5,  and  f(z) = 1.  Then  m = 5 = n + 1 implies that  n = 4.

Remove[z, w]; 
Clear[f, z0]; 
f[z_] = 1; 
n  =  4; 
z0 = (Subscript[z, 0]); 
w[z_]  =  
        (1/(((z - z0))^(n + 1))); 
fn[z_] = D[f[z], {z, n}]; 
Print["f[z] = ", f[z]]; 
Print["n = ", n]; 
\(\(Print[\*""\!\(f\^\((n)\)\)[z] = "", 
        fn[z]];\)\ \)
Print[""]; 
\(\(Print[\*""  \!\(z\_0\) = "", 
        z0];\)\ \)
\(\(Print[\*""\!\(f\^\((n)\)\)["", z0, "] = ", 
        fn[z0]];\)\ \)
Print[""]; 
Print["Use Cauchy's Integral Formulae for Derivatives"]; 
\(\(Print[\*""(∫_C)((f (z))/(((z - (Subscript[z, 0])))^(n + 1)))\[DifferentialD]z = \!\(\(2  π\\ I\)\/\(n!\)\) (f^((n)))["", 
        z0, \*""] = \!\(\(2  π\\ I\)\/\(n!\)\)("", 
        fn[z0], ")"];\)\ \)
\(\(Print[\*""(∫_C)((f (z))/(((z - (Subscript[z, 0])))^(n + 1)))\[DifferentialD]z = "", 
        2\ π\ I\ fn[z0]];\)\ \)

Thus, we have found the value of the contour integral.

dot = 
        Graphics[{Cyan, PointSize[0.03], 
            Point[{2, (5/3)}]}]; 
\(\(center = 
        Graphics[{Text[\*""\!\(z\_0\)"", {2, 1.6}, {0, 
                1}]}];\)\ \)
contour  =  Graphics[{Blue, Circle[{2, (5/3)}, 1]}]; 
Show[center, contour, dot, PlotRange -> {{0, 4}, {0, 3}}, 
        Ticks -> None, 
AspectRatio -> (3/4), 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(Print[\*""\!\(∮\+C\) "", 
      w[z], " \[DifferentialD]z = ", 
      2\ π\ I\ f[z0]]\)
Print["For illustration,"]; 
\(\(Print[\*""a contour could be the unit circle C :|z - (Subscript[z, 0])| = r,""];\)\ \)
Print["with positive orientation."];

Example 6.25,  Page 247.  Show that\(∫\_C\( e\^\(z\^2\)\/\((z\  - \ i)\)\^4\) dz = \(\(-\ 4\)\ π\)\/\(3\ e\)\),  where C is the circle  C : |z| = 2  with positive orientation.

Solution 6.25.

Enter the integrand  w[z] = ((e^((z^2)))/(((z  -  i))^4))  and locate the singularities.

Remove[w, z, Z]; 
Clear[f, i, solset, val]; 
w[
          z_]  =  ((E^((z^2)))/(((z  -  I))^4)); 
solset  =  
        Sort[Solve[Denominator[w[z]] == 0, z]]; 
Z  =  Table[0, {Length[solset]}]; 
For[i = 1, 
        i ≤ 
          Length[solset], i++, 
(Subscript[Z, [[i]]])  =  
            ReplaceAll[z, 
              (Subscript[solset, [[i]]])]; ]; 
Print["w[z] = ", w[z]]; 
Print["The set of singularities is ", 
        Z];

Find the singularity that lie inside  C : |z| = 2.

\(\(\(
\)\(\(Print[\*""(Subscript[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  -  (Subscript[Z, [[1]]])))^4); 
Print["w[z] = ", w[z]]; 
\(\(Print["w[z] = (", 
        f[z], ")/(", \((z\  - \ Z\_\(\([[\)\(1\)\(]]\)\))\)\^4, ")"];\)\ \)
Print[""]; 
Print["f[z] = ", f[z]];

Use Cauchy's Integral Formula for Derivatives to evaluate the integral of  w[z] = ((e^((z^2)))/(((z  -  i))^4)) taken over C.

f3  =  
        Simplify[Expand[f'''[z]]]; 
val  =  ((2 π I)/(3!)) f'''[
            (Subscript[Z, [[1]]])]; 
\(\(Print["w = ", "f[z]"\/\((z\  - \ Z\_\(\([[\)\(1\)\(]]\)\))\)\^4];\)\ \)
Print[""]; 
Print["f[z] = ", f[z]]; 
Print[""]; 
Print[" f'''[z] = ", f3]; 
\(\(Print[\*""      (Subscript[z, 1]) = "", 
        Z\_\(\([[\)\(1\)\(]]\)\)];\)\ \)
\(\(Print[\*""f'''[(Subscript[z, 1])] = "", \ \(\(\(f'\)'\)'\)[
          Z\_\(\([[\)\(1\)\(]]\)\)]];\)\ \)
Print["       n = 3"]; 
Print[""]; 
\(Print["Use Cauchy's Integral Formulae for Derivatives"]; \ 
    Print[\*""(∫_C) "", 
      w[z], \*"" \[DifferentialD]z = (∫_C)((f (z))/(((z - (Subscript[z, 1])))^(n + 1)))\[DifferentialD]z = \!\(\(2  π\\ I\)\/\(n!\)\) (f^((n)))[(Subscript[z, 1])]""]\)
\(\(Print[\*""(∫_C) "", 
        w[z], \*"" \[DifferentialD]z = \!\(\(2  π\\ I\)\/\(3!\)\) f'''["", 
        Z\_\(\([[\)\(1\)\(]]\)\), \*""] = \!\(\(2  π\\ I\)\/\(3!\)\)("", \(f'''\)[
          Z\_\(\([[\)\(1\)\(]]\)\)], ") = ", val];\)\ \)

Thus, we have found the value of the contour integral.

dots = 
        Graphics[{Red, PointSize[0.03], 
            Point[{Re[
                  (Subscript[Z, [[1]]])], Im[(Subscript[Z, [[1]]])]}]}]; 
\(\(centers = 
        Graphics[{Text[\*""(Subscript[z, 1])"", {Re[
                  Z\_\(\([[\)\(1\)\(]]\)\)], Im[Z\_\(\([[\)\(1\)\(]]\)\)] + 
                  0.1}, {\(-1\), 0}]}];\)\ \)
contour  =  Graphics[{Magenta, Circle[{0, 0}, 2]}]; 
Show[dots, centers, contour, 
        PlotRange -> {{-2.1, 2.1}, {-2.1, 2.1}}, 
        Ticks -> {Range[-2, 2, 1], 
            Range[-2, 2, 1]}, 
AspectRatio -> 1, 
        Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(Print[\*""\!\(∮\+C\) ("", 
      w[z], ") \[DifferentialD]z = ", val]\)
Print["The contour is the circle  C: |z| = 2,"]; 
Print["with positive orientation."];

We now state two important corollaries of Theorem 6.12.

Corollary 6.2,  Page 248.  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.

Proof of Corollary 6.1, see text Page 248.

Remark 6.3. This result is interesting, as it illustrates a big difference between real and complex functions. It is possible for a real function f to have the property that f' exists everywhere in a domain D, but f'' exists nowhere. Corollary 6.2 states that if a complex function f has the property that f' exists everywhere in a domain D, then, remarkably, all derivatives of f exist in D.

Corollary 6.3,  Page 248.  If u (x, y) is a harmonic function at each point (x,y) in the domain D, then all partial derivatives  (Subscript[u, x]) (x, y), (Subscript[u, y]) (x, y),  (Subscript[u, xx]) (x, y), (Subscript[u, xy]) (x, y), (Subscript[u, yy]) (x, 
          y),  ...   exists and are harmonic functions.

Proof of Corollary 6.2, see text Page 248.

Library Research Experience for Undergraduates

Project I. Write a report on Cauchy integral formula.

1. Bera, Rajendra K., (1989), ''The Numerical Evaluation of Cauchy Integrals,'' Int. J. of Math. Ed. in Sci. and Tech., V. 20, No. 5, pp. 763-776.

2. Dixon, John D., (1971), ''A Brief Proof of Cauchy's Integral Theorem,'' Proc. Am. Math. Soc., V. 29, No. 3, pp. 625-626.

3. Flatto, L. and O. Shista, (1973), ''A Proof of Cauchy's Integral Theorem,'' J. of Approx. Theory, Vol. 7, pp. 386-390.

4. Konnully, A. O., (1967), ''A Proof of the Cauchy Integral Theorem (in Classroom Notes),'' Am. Math. M., Vol. 74, No. 1, Part 1. (Jan., 1967), pp. 54-56, Jstor.   

5. Kraines, David P., Vivian Y. Kraines and David Smith, (1990), ''The Cauchy Integral Formula,'' Coll. Math. J., V. 21, No. 4, pp. 327-329.

6. Loeb, Peter A., ''A Note on Dixon's Proof of Cauchy's Integral Theorem (in Notes),'' Am. Math. M., Vol. 98, No. 3. (Mar., 1991), pp. 242-244, Jstor.   

7. Loeb, Peter A.,''A Further Simplification of Dixon's Proof of Cauchy's Integral Theorem (in Notes),'' Am. Math. M., Vol. 100, No. 7. (Aug. - Sep., 1993), pp. 680-681, Jstor.   

8. Miller, Norman, ''Discussions: A Corollary to Cauchy's Integral Formula (in Questions and Discussions),'' Am. Math. M., Vol. 35, No. 3. (Mar., 1928), pp. 134-136., Jstor.   

9. Vyborny, R., ''On the Use of a Differentiable Homotopy in the Proof of the Cauchy Theorem (in Classroom Notes),'' Am. Math. M., Vol. 86, No. 5. (May, 1979), pp. 380-382, Jstor.   

10. Walsh, J. L., ''Note on Cauchy's Integral Formula,'' The Annals of Math., 2nd Ser., Vol. 18, No. 2. (Dec., 1916), pp. 79-80., Jstor.   

11. Widder, D. V, ''A Simplified Approach to Cauchy's Integral Theorem,'' Am. Math. M., Vol. 53, No. 7. (Aug. - Sep., 1946), pp. 359-363, Jstor.   

12. Wyler, Oswald, ''The Cauchy Integral Theorem (in Classroom Notes),'' Am. Math. M., Vol. 72, No. 1. (Jan., 1965), pp. 50-53., Jstor.

Section 6.5  Exercises for Integral Representations for Analytic Functions
See textbook page 248.

Section 6.6	The Theorems of Morera and Liouville and Some Applications

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