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 7  TAYLOR AND LAURENT SERIES
Section 7.1  Uniform Convergence
Section 7.2  Taylor series Representations
Section 7.3  Laurent series Representations
Section 7.4  Singularities, Zeros and Poles
Section 7.5  Applications of Taylor and Laurent Series
GoTo Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.
GoTo Contents

CHAPTER 7  TAYLOR AND LAURENT SERIES

Clear[A, f, f1, f2, g, h, i, lim1, lim2, L, L1, L2, L3, L4, m, n, p, 
      solset, S, SI, Sinf, S0, S7, S9, S11, S12, S13, S17, S21, S22, S25, x, 
      y, z, Z1, Z2, Z3, Z4];

Section 7.1  Uniform Convergence

Definition 7.1,  Page 209.  The sequence  {(Subscript[S, n]) (z)}  converges uniformly to  f(z)  on the set  T  if for every  ϵ > 0,  there exists a positive integer  (Subscript[N, ϵ])  (which depends only on ϵ) such that
if  n ≥ N,  then  | (Subscript[S, n]) (z) - f (z) 
        |  <  ϵ  for all  z ϵ T.

Theorem 7.1,  Page 210.  (Weierstrass M-test)  Suppose the infinite series  (∑_(k = 0)^∞)(Subscript[u, k]) (z)  has the property that for each  k,  | (Subscript[u, k]) (z) | 
        (Subscript[( <  M), k])  for all  z ϵ T.   If    (∑_(k = 0)^∞)(Subscript[M, k])  converges,  then  (∑_(k = 0)^∞)(Subscript[u, k]) (z)  converges uniformly on  T.

Theorem 7.2,  Page 211.  Suppose the power series  (∑_(k = 0)^∞)(Subscript[c, k]) (((z - α))^k)  has radius of convergence  ρ  > 0,   Then for each  r,   0 < r <  ρ ,  the series converges uniformly on the closed disk  \(\(\(D\_r\)\&_\) \((α)\) = {
          \(z : \)\  | 
            \(z - α | \( ≤ \ r\)\)}\).

Corollary 7.1,  Page 211.  For each  r,   0 < r <  1  ,  the geometric series converges uniformly on the closed disk  \(\(\(D\_r\)\&_\) \((0)\) = {
          \(z : \)\  | 
            \(z | \( ≤ \ r\)\)}\).

Theorem 7.3,  Page 211.  Suppose   {(Subscript[S, k]) (z)}  is a sequence of continuous functions defined on a set  T  containing the contour  C.   If  {(Subscript[S, k]) (z)}  converges uniformly to  f(z)  on the set  T,  then
(i)	f(z)   is continuous on  T,  and
(ii)	(lim_(k -> ∞))(∫_C^( )) (Subscript[S, k])(z)dz = (∫_C^( )) (lim_(k -> ∞))(Subscript[S, k])(z)dz =(∫_C^( )) f(z)dz .

Corollary 7.2,  Page 212.  If the series  (∑_(k = 0)^∞)(Subscript[c, k]) (((z - α))^k)  converges uniformly to  f(z)  on the set  T,  and  C  is a contour contained in  T,  then
  (∑_(k = 0)^∞)(∫_C^( )) (Subscript[c, k])(((z - α))^k)dz = (∫_C^( )) (∑_(k = 0)^∞)(Subscript[c, k])(((z - α))^k)dz  =(∫_C^( )) f(z)dz .

Load the SymbolicSum package.  Make sure this is done only once in a Mathematica  session.

Example 7.1,  Page 209.  The sequence  {(Subscript[S, n]) (z)} =  {(e^z)  + (1/n)}  converges uniformly to   (e^z)  on the entire complex plane.
Solution.  The solution is given in the textbook.

Example 7.2,  Page 212.  Show that   Log(1 - z)  = - (∑_(n = 1)^∞)(((z^n))/n)   for  all   z ∈ D.

\(Clear[f, n, S17, S, Sinf, z]; 
f[z_]\  = \ Log[1 - z]; 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 17}]]; 

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

    Print[\*""\!\(S\_17\)[z] = "", S[z]]; \)

Sum up the terms to verify that we have things right.

\(S17[z_]\ \ \  = \ ∑\+\(n = 1\)\%17\ \(-\ z\^n\)\/n; 

    Sinf[z_]\  = \ ∑\+\(n = 1\)\%∞\ \(-\ z\^n\)\/n; 

    Print[\*""(Subscript[S, 17])[z] = "", S17[z]]; 

    Print[\*""(Subscript[S, ∞])[z] = (∑_(n = 1)^∞)\!\(\(-\\ z\^n\)\/n\) = "", Sinf[z]]; 
Print[" f[z] = ", f[z]]; 
    \)

Plot[{f[x], S17[x]}, {x, -0.99999, 0.99999}, 
      PlotRange -> {{-1, 1}, {-6, 1}}, 
      Ticks -> {Range[-1, 1, 0.5], Range[-6, 1, 1]}, 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, 
      AxesLabel -> {"x", "s"}];

GoTo Chapter 7

Section 7.2  Taylor series Representations

In Section 4.2  we saw that functions defined by power series have derivatives of all orders.  In Section 6.5 we saw that analytic functions also have derivatives of all orders.  It seems natural, therefore, that there would be some connection between analytic functions and power series.  As you might guess, the connection exists via the Taylor and Maclaurin series of analytic functions.

Definition 7.2,  Page 214.  If  f (z)  is analytic at  z = α,  the series  (∑_(k = 0)^∞)((Derivative[k][f](α))/(k!))(((z - α))^k)  is called the Taylor series for  f (z)  centered around  α.  When the center is  α = 0,  the series is called the Maclaurin series for  f (z).
  
Theorem 7.4,  Page 215.  (Taylor's Theorem)  Suppose  f (z)  is analytic in a domain  G,  and that  (Subscript[D, R]) (α) = {
          z :   | 
            z - α |   <  R}  is contained in  G.   Then the Taylor series converges to  f (z)  for all  z  in  (Subscript[D, R]) (α);  that is  f (z) = (∑_(k = 0)^∞)((Derivative[k][f](α))/(k!))(((z - α))^k)  for all  z ϵ (Subscript[D, R]) (α).
     Furthermore, this representation is valid in the largest disk with center  z = α  that is contained in  G,  and the convergence is uniform on any closed subdisk  \(\(\(D\_r\)\&_\) \((α)\) = {
          \(z : \)\  | 
            \(z - α | \( ≤ \ r\)\)}\)  for  0 < r <  R  . 

Corollary 7.5,  Page 217.  Suppose  f (z)  is analytic in the domain  G  that contains the point  z = α.   Let  (Subscript[z, 0])  be a singular point of minimum distance to  α.   If  | (Subscript[z, 0]) - α |  = 
         R,  then
(i)	the Taylor series converges to  f (z)  on all of  (Subscript[D, R]) (α),  and
(ii)	if  S > R,  the Taylor series does not converge to  f (z)  on all of   (Subscript[D, S]) (α).

Theorem 7.5,  Page 218.  (Uniqueness of Power Series)  Suppose that in some disk  (Subscript[D, r]) (α)  we have  f (z) =  (∑_(n = 0)^∞)(Subscript[a, n]) (((z - α))^n) =  (∑_(n = 0)^∞)(Subscript[b, n]) (((z - α))^n) .   
Then  (Subscript[a, n]) = (Subscript[b, n])  for   n = 0, 1, 2, ⋯.

Theorem 7.6,  Page 219.  Let f (z) and g (z) have the power series representations
f (z) =  (∑_(n = 0)^∞)(Subscript[a, n]) (((z - α))^n)   for   z ϵ 
            (Subscript[D, (Subscript[r, 1])]) (α)  andg (z) =  (∑_(n = 0)^∞)(Subscript[b, n]) (((z - α))^n)   for   z ϵ 
            (Subscript[D, (Subscript[r, 2])]) (α).
If   r  =  min {(Subscript[r, 1]), (Subscript[r, 2])},  and  β  is any complex constant, then
(i)	βf (z) =  (∑_(n = 0)^∞)β(Subscript[a, n]) (((z - α))^n)   for   z 
            ϵ (Subscript[D, (Subscript[r, 1])]) (α),

(ii)	f (z) + g (z) =  (∑_(n = 0)^∞)((Subscript[a, n]) + (Subscript[b, n])) (((z - α))^n)   for   z 
            ϵ (Subscript[D, r]) (α),  and

(iii)	f (z)*g (z) =  (∑_(n = 0)^∞)(Subscript[c, n]) (((z - α))^n)   for   z ϵ 
            (Subscript[D, r]) (α),  where
	(Subscript[c, n])  =  (∑_(k = 0)^n)(Subscript[a, k]) (Subscript[b, n - k]).
Identity (iii) is known as the Cauchy product of the series for  f (z) and g (z).

Example 7.3,  Page 217.  Show that   (1/((((1  -  z))^2))) = (∑_(n = 0)^∞)(n + 1) (z^( n))  is valid for  z ∈ D.

\(Clear[f, n, S, S17, Sinf, z]; 
f[z_]\  = \ 1\/\((1 - z)\)\^2; 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 17}]]; 

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

    Print[\*""\!\(S\_17\)[z] = "", S[z]]; \)

Sum up the terms to verify that we have things right.

\(S17[z_]\ \ \  = \ ∑\+\(n = 0\)\%17\((n + 1)\)\ z\^n; 

    Sinf[z_]\  = \ ∑\+\(n = 0\)\%∞\((n + 1)\)\ z\^n; 

    Print[\*""(Subscript[S, 17])[z] = "", S17[z]]; 

    Print[\*""(Subscript[S, ∞])[z] = (∑_(n = 0)^∞)(n+1) (z^n) = "", Sinf[z]]; \)

Example 7.4,  Page 218.  (a)  Show that   (1/(1 - (z^2)))  =  (∑_(n = 0)^∞) (z^( 2 n))  is valid for  z ∈ D.

\(Clear[f, n, S25, S, Sinf, z]; 
f[z_]\  = \ 1\/\(1 - z\^2\); 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 50}]]; 

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

    Print[\*""\!\(S\_50\)[z] = "", S[z]]; \)

Sum up the terms to verify that we have things right.

\(S25[z_]\ \ \  = \ ∑\+\(n = 0\)\%25 z\^\(2  n\); 

    Sinf[z_]\  = \ \ ∑\+\(n = 0\)\%∞ z\^\(2  n\); 

    Print[\*""(Subscript[S, 25])[z] = "", S25[z]]; 

    Print[\*""(Subscript[S, ∞])[z] = (∑_(n = 0)^∞)(z^(2  n)) = "", Sinf[z]]; \)

(b)  Show that   (1/(1 + (z^2))) = (∑_(n = 0)^∞) (((-1))^n) (z^( 2 n))   is valid for  z ∈ D.

\(Clear[f, n, S, S25, Sinf, z]; 
f[z_]\  = \ 1\/\(1 + z\^2\); 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 50}]]; 

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

    Print[\*""\!\(S\_50\)[z] = "", S[z]]; \)

Sum up the terms to verify that we have things right.

\(S25[z_]\ \ \  = \ ∑\+\(n = 0\)\%25\(\((\(-1\))\)\^n\) z\^\(2  n\); 
    
Sinf[z_]\  = \ \ 
      ∑\+\(n = 0\)\%∞\(\((\(-1\))\)\^n\) z\^\(2  n\); 

    Print[\*""(Subscript[S, 25])[z] = "", S25[z]]; 

    Print[\*""(Subscript[S, ∞])[z] = (∑_(n = 0)^∞)(((-1))^n)(z^(2  n)) =  "", Sinf[z]]; \)

Example 7.5,  Page 219.  Find the Maclaurin series of   f(z) = (sin^3)(z).

\(Clear[f, f1, f2, n, S, z]; 
f[z_]\  = \ Sin[z]\^3; 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 25}]]; 

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

    Print[\*""\!\(S\_25\)[z] = "", S[z]]; \)

Sum up the terms to verify that we have things right.

\(f1[z_]\  = 
      PowerExpand[
        ∑\+\(n = 0
              \)\%∞\(\((\(-1\))\)\^n\ 3\)\/\(4 
                \(\((2  n + 1)\)!\)\)\ z\^\(2  n + 1\)]; 

    f2[z_]\  = \ 
      PowerExpand[
        ∑\+\(n = 0
              \)\%∞\(\(-\ \((\(-1\))\)\^n\)\ 3\ 9\^n\)\/\(4 
                \(\((2  n + 1)\)!\)\)\ z\^\(2  n + 1\)]; 

    f[z_]\  = \ Factor[f1[z]\  + \ f2[z], \ Trig -> True]; 

    Print[\*""(Subscript[f, 1])[z] = (∑_(n = 0)^∞)\!\(\(\((\(-1\))\)\^n\\ 3\)\/\(4 \(\((2  n + 1)\)!\)\)\) (z^(2  n + 1)) =  "", f1[z]]; 

    Print[\*""(Subscript[f, 2])[z] = (∑_(n = 0)^∞)\!\(\(\(-\\ \((\(-1\))\)\^n\)\\ 3\\ 9\^n\)\/\(4 \(\((2  n + 1)\)!\)\)\) (z^(2  n + 1)) =  "", f2[z]]; 

    Print[\*"" f[z] = (Subscript[f, 1])[z] + (Subscript[f, 2])[z] = "", \ 
      f1[z]\  + \ f2[z]]; 
Print[" f[z] = ", f[z]]; \)

Example 7.6,  Page 220.  Use the Cauchy product of series to show that  
(1/((((1  -  z))^2)))=(∑_(n = 0)^∞)(n + 1) (z^n)   for  z ϵ (Subscript[D, 1]) 
            (0).
Solution.  The solution is given in the textbook.

GoTo Chapter 7

Section 7.3  Laurent series Representations

The main result of this section specifies how functions analytic in an annulus can be expanded in a Laurent series.  Often times this just means including terms with negative powers of  z .

Theorem 7.7,  Page 225.  Suppose the Laurent series  (∑_(n=-∞)^∞)(Subscript[c, n]) (((z - α))^n)   converges on an annulus    A (r, R, α) = {
          z :  r  <  | 
            z - α |  <  R}.  Then the series converges uniformly on any subannulus  \(\(A\&_\) \((s, t, α)\) = {
          z : \ \(s\  ≤ \) | 
            \(z - α | \( ≤ \ t\)\)}\)  where  r  < s < t <  R.

Theorem 7.8,  Page 225.  (Laurent's Theorem)  Suppose  0 ≤ r < R,  and that  f (z)  is analytic in the annulus  A = 
        A (r, R, α) = {
            z :  r  <  | 
              z - α |  <  R}.   If  ρ  is any number such that  r  < ρ <  R,  then for all  (Subscript[z, 0]) ϵ A (r, R, α),   f (z)  has the Laurent series representation 
f ((Subscript[z, 0]))=(∑_(n=-∞)^∞)(Subscript[c, n]) ((((Subscript[z, 0]) - α))^n)=(∑_(n = 1)^∞)(Subscript[c, -n]) ((((Subscript[z, 0]) - α))^(-n))+(∑_(n = 0)^∞)(Subscript[c, n]) ((((Subscript[z, 0]) - α))^n)
where for  n = 0, 1, 2, ⋯  the coefficients (Subscript[c, n])  and  (Subscript[c, -n])  are given by
(Subscript[c, -n])=(1/(2 π i))(∫_((Subscript[(SuperPlus[C]), ρ])(α))^( )) ((f(z))/(((z - α))^(-n + 1)))dz   and   (Subscript[c, n])=(1/(2 π i))(∫_((Subscript[(SuperPlus[C]), ρ])(α))^( )) ((f(z))/(((z - α))^(n + 1)))dz.

Theorem 7.9,  Page 228.  (Uniqueness and differentiation of Laurent Series)  Suppose that  f (z)  is analytic in the annulus  A (r, R, α),  and has the Laurent series representation  f (z)=(∑_(n=-∞)^∞)(Subscript[c, n]) (((z - α))^n)   for all   z ϵ A
             (r, R, α).
(i)	If  f (z)=(∑_(n=-∞)^∞)(Subscript[b, n]) (((z - α))^n)   for all   z ϵ A
             (r, R, α),  then  (Subscript[b, n]) = (Subscript[c, n])  for all  n.  (In other words, the Laurent series for  f (z)  in a given annulus is unique.)
(ii)	For all  z ϵ A (r, R, α),  the derivatives for  f (z)  may be obtained by termwise differentiation of its Laurent series.

Load the SymbolicSum package.  Make sure this is done only once in a Mathematica  session.

Example on,  Page 224.  Find the Laurent series for   f(z)  =  (((e^z))/((z^3))).

\(Clear[f, L, S, S9, S11, S12, S21, S22, Sinf, z]; 

    f[z_]\  = \ E\^z\/z\^3; 

    S[z_]\  = \ Normal[Series[E\^z, {z, 0, 9}]]\/z\^3; 

    L[z_]\  = \ Normal[Series[f[z], {z, 0, 6}]]; 

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

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

The latter is a Laurent series is valid for 0 < |z|.

Sum up the terms to verify that we have things right.

\(S9[z_]\  = \ ∑\+\(n = 0\)\%9 z\^\(n - 3\)\/\(n!\); 

    Sinf[z_]\  = \ \ ∑\+\(n = 0\)\%∞ z\^\(n - 3\)\/\(n!\); 

    Print[\*""(Subscript[S, 9])[z] = "", S9[z]]; 

    Print[\*""(Subscript[S, ∞])[z] = (∑_(n = 0)^∞)((z^(n - 3))/(n!)) =  "", Sinf[z]]; \)

\(Print["s = ", f[x], \*""  and  s = \!\(S\_9\)[x]""]; 

    Plot[{f[x], S9[x]}, {x, 0.1, 8}, PlotRange -> {{0, 8}, {0, 10}}, 
      Ticks -> {Range[0, 8, 1], Range[0, 10, 1]}, 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, 
      AxesLabel -> {"x", "s"}]; \)

Example 7.7,  Page 229.  Find three different Laurent series representations for
f(z)  =  (3/(2 + z - (z^2)))  involving powers of  z.

\(Clear[A, f, f1, f2, L1, L2, L3, L4, S0, S11, S12, S21, S22, SI, z]; 

    f[z_]\  = \ 3\/\(2\  + \ z\  - \ z\^2\); 
Print["f[z] = ", f[z]]; 
    \)

Mathematica  easily finds the Maclaurin series involving positive powers of z:

\(S0[z_]\  = \ Normal[Series[f[z], {z, 0, 7}]]; 

    Print[\*""\!\(S\_0\)[z] = "", S0[z]]; \)

The above series converges for |z| < 1, this requires looking at the sum of the two 
"geometric series that form the parts of S0."

Mathematica  can also find "the other series" that involves negative powers of  z, this involves the mental thinking that you substitute  Z = (1/z)  in the original series to get a new function of  Z, then expand it about  Z = 0  and then substitute  Z = (1/z).

\(SI[z_]\  = \ Normal[Series[f[z], {z, ∞, 10}]]; 

    Print[\*""\!\(S\_I\)[z] = "", SI[z]]; \)

The above series converges for 2 < |z|, this requires looking at the sum of the two "geometric series that form the parts of (Subscript[S, I]).

This leaves an unresolved question: "What happens in the annulus 1 < |z| < 2 ? 
This requires a little work.  First split up the functions up into their partial fraction form, 
and then do expansions about z=0 and z=∞ for each part.

\(A\  = \ MapAll[Cancel, Apart[f[z]]]; 

    f1[z_]\  = \ A[[2]]; 

    f2[z_]\  = \ A[[1]]; 

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

    Print[\*""\!\(f\_1\)[z] = "", f1[z]]; 

    Print[\*""\!\(f\_2\)[z] = "", f2[z]]; \)

Now form Laurent expansions for the two parts  (Subscript[f, 1])(z)  and  (Subscript[f, 2])(z)  of  f(z).

\(S11[z_]\  = \ Normal[Series[f1[z], {z, 0, 12}]]; 

    S12[z_]\  = \ Normal[Series[f1[z], {z, ∞, 11}]]; 

    S21[z_]\  = \ Normal[Series[f2[z], {z, 0, 12}]]; 

    S22[z_]\  = \ Normal[Series[f2[z], {z, ∞, 12}]]; 

    Print[\*""\!\(S\_11\)[z] = "", S11[z]]; 

    Print[\*""\!\(S\_12\)[z] = "", S12[z]]; 

    Print[\*""\!\(S\_21\)[z] = "", S21[z]]; 

    Print[\*""\!\(S\_22\)[z] = "", S22[z]]; \)

The Laurent series  L1(z)  =  S11(z) + S21(z)  is valid for |z| < 1.

\(L1[z_]\  = \ S11[z]\  + \ S21[z]; 

    Print[\*""\!\(L\_1\)[z] = "", L1[z]]; \)

\(Print["s = ", f[x], \*""  and  s = \!\(L\_1\)[z]""]; 

    Plot[{f[x], L1[x]}, {x, 0.0, 0.999999}, 
      PlotRange -> {{0, 1}, {1, 2}}, 
      Ticks -> {Range[0, 1, 0.2], Range[1, 2, 0.2]}, 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, 
      AxesLabel -> {"x", "s"}]; \)

The following Laurent series  L2(z)  =  S12(z) + S21(z)  is valid for 1 < |z| < 2.

\(L2[z_]\  = \ S12[z]\  + \ S21[z]; 

    Print[\*""\!\(L\_2\)[z] = "", L2[z]]; \)

\(Print["s = ", f[x], \*""  and  s = \!\(L\_2\)[x]""]; 

    Plot[{f[x], L2[x]}, {x, 1.0001, 1.9999}, 
\ \ 
      PlotRange -> {{1, 2}, {0, 7}}, 
      Ticks -> {Range[1, 2, 0.2], Range[0, 7, 1]}, 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, 
      AxesLabel -> {"x", "s"}]; \)

So L2(z) is an approximation to f[z] valid for 1 < |z| < 2.
Notice that L1(z) is not a good approximation for 1 < |z| < 2,
for example it is not accurate at  z = 1.5

\(Print[" f[1.5] = ", f[1.5]]; 

    Print[\*""\!\(L\_1\)[1.5] = "", L1[1.5]]; 

    Print[" f[1.5] = ", f[1.5]]; 

    Print[\*""\!\(L\_2\)[1.5] = "", L2[1.5]]; \)

The Laurent series  L3(z)  =  S12(z) + S22(z)  is valid for 2 < |z| .

\(L3[z_]\  = \ S12[z]\  + \ S22[z]; 

    Print[\*""\!\(L\_3\)[z] = "", L3[z]]; \)

\(Print["s = ", f[x], \*""  and  s = \!\(L\_3\)[x]""]; 

    Plot[{f[x], L3[x]}, {x, 2.0001, 4.0}, 
\ \ 
      PlotRange -> {{2, 4}, {\(-6\), 0}}, 
      Ticks -> {Range[2, 4, 0.2], Range[\(-6\), 0, 1]}, 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, 
      AxesLabel -> {"x", "s"}]; \)

So L3(z) is an approximation to f[z] valid for 2< |z|.
Notice that L1(z) and L2(z)  are not good approximations for 2< |z|,
for example they are not accurate at  z = 4.0

\(Print[" f[4.0] = ", f[4.0]]; 

    Print[\*""\!\(L\_1\)[4.0] = "", L1[4.0]]; 

    Print[" f[4.0] = ", f[4.0]]; 

    Print[\*""\!\(L\_2\)[4.0] = "", L2[4.0]]; 

    Print[" f[4.0] = ", f[4.0]]; 

    Print[\*""\!\(L\_3\)[4.0] = "", L3[4.0]]; \)

The fourth possibility does not converge for any values of z and should not be considered, however you could write down a few terms and think about why this series diverges.

\(L4[z_]\  = \ S11[z]\  + \ S22[z]; 

    Print[\*""\!\(L\_4\)[z] = "", L4[z]]; \)

In the above series, when  |z| < (1/2)  the portion of the series with positive exponents will
converge but the portion with negative exponents will diverge. Similarly, when |z| > 1 the 
portion of the series with negative exponents will converge, but the portion of the series
with positive exponents will diverge.   Hence, the series diverges for all  z.

\(Print[" f[0.5] = ", f[0.5]]; 

    Print[\*""\!\(L\_4\)[0.5] = "", L4[0.5]]; 

    Print[" f[1.5] = ", f[1.5]]; 

    Print[\*""\!\(L\_4\)[1.5] = "", L4[1.5]]; 

    Print[" f[4.0] = ", f[4.0]]; 

    Print[\*""\!\(L\_4\)[4.0] = "", L4[4.0]]; \)

Example 7.8,  Page 229.  Find the Laurent series for   f(z) = ((cos (z)  -  1)/((z^4))).

\(Clear[f, L, S, S7, Sinf, z]; 
f[z_]\  = \ \(Cos[z]\  - \ 1\)\/z\^4; 

    S[z_]\  = \ Normal[Series[Cos[z] - 1, \ {z, 0, 14}]]\/z\^4; 

    L[z_]\  = \ Normal[Series[f[z], {z, 0, 10}]]; 

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

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

The latter Laurent series is valid for  0 < |z| .

Sum up the terms to verify that we have things right.

\(S7[z_]\  = \ 
      ∑\+\(n = 1\)\%7
          \(\(\((\(-1\))\)\^n\) z\^\(2  n - 4\)\)\/\(\((2  n)\)!\); 

    Sinf[z_]\  = \ \ 
      ∑\+\(n = 1\)\%∞\(\(\((\(-1\))\)\^n\) 
            z\^\(2  n - 4\)\)\/\(\((2  n)\)!\); 

    Print[\*""(Subscript[S, 7])[z] = "", S7[z]]; 

    Print[\*""(Subscript[S, ∞])[z] = (∑_(n = 1)^∞)(((((-1))^n) (z^(2  n - 4)))/((2  n)!)) =  "", 
      Sinf[z]]; 

    Print[\*""(Subscript[S, ∞])[z] = (∑_(n = 1)^∞)(((((-1))^n) (z^(2  n - 4)))/((2  n)!)) =  "", 
      PowerExpand[Sinf[z]]]; \)

\(Print["f[0.2]= ", f[0.2]]; 
Print["L[0.2]= ", L[0.2]]; 

    Print["f[2.0]= ", f[2.0]]; 
Print["L[2.0]= ", L[2.0]]; \)

\(Print["s = ", f[x], "  and  s = L[x]"]; 

    Plot[{f[x], L[x]}, {x, 0.0001, 5.0}, 
\ \ 
      PlotRange -> {{0, 5}, {\(-5\), 0.1}}, 
      Ticks -> {Range[0, 5, 1], Range[\(-5\), 0, 1]}, 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, 
      AxesLabel -> {"x", "s"}]; \)

Example 7.9,  Page 230.  Find the Laurent series for  f(z) = (e^(((-1)/(z^2)))) 
centered at  (Subscript[z, 0]) = 0.

\(Clear[f, L, L1, S1, z]; 

    f[z_]\  = \ E\^\(\(-1\)\/z\^2\); 

    L[z_]\  = \ Normal[Series[f[z], {z, ∞, 20}]]; 

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

Note that we had to expand the function about  ∞  in order to get the negative powers of  z.
Also, can get the Laurent expansion by series substitution:

\(S1[z_]\  = \ Normal[Series[E\^z, {z, 0, 10}]]; 

    L1[z_]\  = \ S1[\(-1\)\/z\^2]; 

    Print[\*""   \!\(E\^z\) = "", S1[z]]; 

    Print[\*""\!\(L\_1\)[z] = "", L1[z]]; \)

The above Laurent series are valid for  0 < |z| .

\(Print["s = ", f[x], \*""  and  s = \!\(L\_1\)[x]""]; 

    Plot[{f[x], L1[x]}, {x, 0.3, 2.0}, 
\ \ 
      PlotRange -> {{0, 2}, {\(-0.1\), 1}}, 
      Ticks -> {Range[0, 2, 0.5], Range[0, 1, 0.2]}, 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, 
      AxesLabel -> {"x", "s"}]; \)

GoTo Chapter 7

Section 7.4  Singularities, Zeros and Poles

What is a singular point ?
     The point  α  is called a singular point, or singularity, of the complex function  f(z)
if  f(z)  is not analytic at  z = α, but every neighborhood  DR(α)  of  α  contains
at least one point at which  f(z)  is analytic.

Theorem 7.10,  Page 234.  A function  f (z) analytic in   (Subscript[D, R]) (α)  has a zero of order  k  at the point  z = α  if and only if its Taylor series given by   f (z) =  (∑_(n = 0)^∞)(Subscript[c, n]) (((z - α))^n)  has  (Subscript[c, 0]) = 
        (Subscript[c, 1]) = 
          ⋯ = 
            (Subscript[c, k - 1]) = 0   and   (Subscript[c, k]) ≠ 0.

Theorem 7.11,  Page 235.  Suppose  f (z)  is analytic in   (Subscript[D, R]) (α).  Then  f (z)  has a zero of order  k  at the point  z = α  if and only if it can be expressed in the form  f (z)  = (((z - α))^k) g (z),  where  g (z)  is analytic at  z = α  and  g (α) ≠ 0.

Theorem 7.12,  Page 236.  A function  f (z)  analytic in the punctured disk  \(\(\ \(D\_R\^*\) \((α)\)\)\)  has a pole of order  k  at  z = α  if and only if it can be expressed in the form   f (z) =((h (z))/(((z - α))^k)),  where  h (z)  is analytic at  z = α  and  h (α) ≠ 0.

Theorem 7.13,  Page 236.  
(i)	If  f (z) is analytic and has a zero of order  k  at  z = α,  
	then  g (z) =(1/(f (z)))  has a pole of order  k  at  z = α.
(ii)	If  f (z) has a pole of order  k  at  z = α,  
	then  h (z) =(1/(f (z)))  has a removable singularity at  z = α.  
	If we define  h (α) = 0,  then  h (z)  has a zero of order  k  at  z = α.

What is a pole ?
     The following example will help this concept. 
Consider the function  f(z) = cot(z).  The leading term in the series expansion  
S(z)  is (1/z)  and  S(z)  goes to  ∞  as  z->0  in the same manner as  cot(z).

\(Clear[f, S, z]; 
f[z_]\  = \ Cot[z]; 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 13}]]; 

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

\(Print["The function  f[x] = ", f[x]]; 

    Print["and series    S[x] = ", S[x]]; 

    graph1\  = \ 
      Plot[{f[z], S[z]}, {z, 0.01, 3.1}, 
        PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, 
        DisplayFunction -> Identity]; \ 

    graph2\  = \ 
      Plot[{f[z], S[z]}, {z, \(-3.1\), \(-0.01\)}, 
        PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, 
        DisplayFunction -> Identity]; \ 

    Show[graph1, graph2, PlotRange -> {{\(-3.1\), 3.1}, {\(-10\), 10}}, 
      Ticks -> {Range[\(-3\), 3, 1], Range[\(-10\), 10, 5]}, 
      DisplayFunction -> $DisplayFunction]; \)

Example of a removable singularity,  Page 233.
The function   f(z) = ((sin(z))/z)   has a removable singularity at  z = 0.

\(Clear[f, S, z]; 
f[z_]\  = \ Sin[z]\/z; 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 13}]]; 

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

    Print["       f[0] = ", f[0]]; 

    Print[\*""\!\(lim\+\(x -> 0\)\) "", f[x], " = ", 
      Limit[f[x], x -> 0]]; 
Print["       S[0] = ", S[0]]; \)

Example of a removable singularity,  Page 233.
The function   f(z) = ((cos(z)  -  1)/((z^2)))   has a removable singularity at  z = 0.

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

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 11}]]; 

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

    Print["            f[0] = ", f[0]]; 

    Print[\*""\!\(lim\+\(x -> 0\)\) "", f[x], " = ", 
      Limit[f[x], x -> 0]]; 
Print["            S[0] = ", S[0]]; 
    \)

Example of a pole of order 2,  Page 234.
The function   f(z) = ((sin(z))/((z^3)))  has a pole of order  2  at the origin.

\(Clear[f, S, z]; 
f[z_]\  = \ Sin[z]\/z\^3; 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 11}]]; 

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

Example of a simple pole,  Page 234.
The function   f(z) = (((e^z))/z)  has a simple pole at the origin.

\(Clear[f, S, z]; 
f[z_]\  = \ E\^z\/z; 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 8}]]; 

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

Examples of an essential singularity,  Page 234.
The function   f(z) = (z^2)sin((1/z))  has an essential singularity at the origin.

\(Clear[g, h, p, S, z]; 
g[z_]\  = \ Sin[z]; 

    S[z_]\  = \ Normal[Series[Sin[z], {z, 0, 17}]]; 
h[z_]\  = \ S[1\/z]; 

    f[z_]\  = \ z\^2\ g[1\/z]; 
p[z_]\  = \ Simplify[z\^2\ h[z]]; 

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

Example 7.10,  Page 234. Show that  f(z) = z sin((z^2))  has a zero of order  3  at  z = 0.

\(Clear[f, i, L, m, p, S, z]; 
f[z_]\  = \ z\ Sin[z\^2]; 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 23}]]; 

    L\  = \ CoefficientList[S[z], z]; 
i = 1; \ m\  = \ Length[L]; 

    While[L[\([i]\)] == 0\  && \ i < m, 
\ \ term\  = \ z\^i; \ i = i + 1]; 
    \ 
fact\  = \ Simplify[S[z]\/term]; 
p[z_]\  = \ term\ fact; 

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

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

Thus,  f(z)  =  z sin((z^2))   has a zero of order  3  at  z = 0.

Example 7.11,  Page 236. Show that  f(z) = (z^3)sin(z)  has a zero of order  4  at  z = 0.

\(Clear[f, i, L, m, p, S, z]; 
f[z_]\  = \ z\^3\ Sin[z]; 

    S[z_]\  = \ Normal[Series[f[z], {z, 0, 16}]]; 

    L\  = \ CoefficientList[S[z], z]; 
i = 1; \ m\  = \ Length[L]; 

    While[L[[i]] == 0\  && \ i < m, 

      \ \ \ term\  = \ z\^i; \ i = i + 1]; \ 

    fact\  = \ Simplify[S[z]\/term\ ]; 
p[z_]\  = \ term\ fact; 

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

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

Thus,  f(z)  =  (z^3)sin(z)   has a zero of order  4  at  z = 0.

Example 7.12,  Page 237.  Locate the zeros and poles of  h(z) = ((tan(z))/z),
and determine their order.

\(Clear[h, i, S, z]; 
h[z_]\  = \ Tan[z]\/z; 

    S[z_]\  = \ Normal[Series[h[z], {z, 0, 13}]]; 

    Print["h[z] = ", h[z]]; 
Print["S[z] = ", S[z]]; \)

Thus,  h(z)  =  ((tan (z))/z)   has a removable singularity at  z = 0.

\(Print[" h[z] = ", h[z]]; 

    For[i = 1, i ≤ 5, \(i++\), 
\ \ \ 
      Print["h[", i, "π] = ", h[i\ π]]\ ]; \)

And   h(z)  =  ((tan (z))/z)   has simple zeros at  z = nπ  where  n = ±1, ±2, ...

\(Print["    h[z] = ", h[z]]; 

    For[i = 1, i <= 5, \(i++\), 
      Print["h[", 2  i - 1, "/2 π] = ", 
        Limit[h[z], z -> \(\((2  i - 1)\) π\)\/2]\ ]\ ]; \)

And   h(z)  =  ((tan (z))/z)   has simple poles at  z = (n+(1/2))π  where n is an integer.

Example 7.13,  Page 237.  Locate the poles of  g(z) = (1/(5(z^4)+ 26(z^2) + 5)),
and specify their order.

\(Clear[f, zZ1, Z2, Z3, Z4]; 

    g[z_]\  = \ 1\/\(5  z\^4\  + \ 26  z\^2\  + \ 5\); 

    solset\  = \ Solve[Denominator[g[z]] == 0, z]; 

    Print["g[z] = ", g[z]]; 
Print["The poles are located at:"]; 

    Print[solset]; \)

\(Z1\  = \ solset[[1, 1, 2]]; 

    Z2\  = \ solset[[2, 1, 2]]; 

    Z3\  = \ solset[[3, 1, 2]]; 

    Z4\  = \ solset[[4, 1, 2]]; 

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

    Print["g[", Z1, "] = ", Limit[g[z], z -> Z1]]; 

    Print["  g[", Z2, "] = ", Limit[g[z], z -> Z2]]; 

    Print["g[", Z3, "] = ", Limit[g[z], z -> Z3]]; 

    Print["  g[", Z4, "] = ", Limit[g[z], z -> Z4]]; \)

Thus  g(z) = (1/(5(z^4)+ 26(z^2) + 5)), has four simple poles.

Example 7.14,  Page 237.  Locate the zeros and poles of  g(z) = ((π cot(πz))/((z^2))),
and determine their order.

\(Clear[g, S, z]; 
g[z_]\  = \ \(π\ Cot[π\ z]\)\/z\^2; 

    S[z_]\  = \ Normal[Series[g[z], {z, 0, 9}]]; 

    Print["g[z] = ", g[z]]; 
Print["S[z] = ", S[z]]; \)

Thus  g(z)  =  ((π cot(πz))/((z^2))), has a pole of order  n = 3  at  z = 0.

\(Print["g[z] = ", g[z]]; 

    Print["g[1] = ", Limit[g[z], z -> 1]]; 

    Print["g[2] = ", Limit[g[z], z -> 2]]; 

    Print["g[3] = ", Limit[g[z], z -> 3]]; 

    Print["g[4] = ", Limit[g[z], z -> 4]]; \)

Thus   g(z)  =  ((π cot(πz))/((z^2)))   has simple poles at  z = ±1, ±2, ...

GoTo Chapter 7

Section 7.5  Applications of Taylor and Laurent Series

In this section we show how Taylor and Laurent series can be used to derive important properties of analytic functions.

Theorem 7.14,  Page 239.  Suppose  f (z)  is analytic at  z = α  and that  f (α) = 0.   If  f (z)  is not identically zero, then there exists a punctured disk  \(\(\ \(D\_r\^*\) \((α)\)\)\)  in which  f (z)  has no zeros.

Corollary 7.9,  Page 239.  (L'Hopital's Rule)  Suppose f (z) and g (z) are analytic at  z = α.   If  f (α) = 0  and  g (α) = 0  but  g'(α)≠0,  then 
(lim_(z -> α))((f (z))/( g (z))) = (lim_(z -> α))((f'(z))/( g'(z))) = ((f'(α))/(g'(α))).  

Theorem 7.15,  Page 240.  (Division of Power Series)  Suppose f (z) and g (z) are analytic at  z = α  with power series representations
f (z) =  (∑_(n = 0)^∞)(Subscript[a, n]) (((z - α))^n)   for   z ϵ 
            (Subscript[D, R]) (α)  andg (z) =  (∑_(n = 0)^∞)(Subscript[b, n]) (((z - α))^n)   for   z ϵ 
            (Subscript[D, R]) (α).
If   g (α) ≠ 0,  then the quotient  ((f (z))/( g (z)))  has the power series representation
((f (z))/( g (z))) =  (∑_(n = 0)^∞)(Subscript[c, n]) (((z - α))^n) ,  
where the coefficients satisfy the equations  (Subscript[a, n]) = 
        (Subscript[b, 0]) (Subscript[c, n]) + (Subscript[b, 1]) (Subscript[c, n - 1]) + ⋯ + 
          (Subscript[b, n - 1]) (Subscript[c, 1]) + (Subscript[b, n]) (Subscript[c, 0]).
In other words, the series for the quotient  ((f (z))/( g (z)))  can be obtained by the familiar process of dividing the series for  f (z)  by the series for   g (z)  using the standard long division algorithm.

Theorem 7.16,  Page 241.  (Riemann's Theorem)  Suppose that  f (z)  is analytic in  \(\(\ \(D\_r\^*\) \((α)\)\)\).   If   f (z)  is bounded in  \(\(\ \(D\_r\^*\) \((α)\)\)\),  then either   f (z)  is analytic at  z = α  or  f (z)  has a removable singularity at  z = α.

Theorem 7.17,  Page 241.  The function  f (z)  has a pole of order  k  at  z = α  if and only if   (lim_(z -> α))| 
          f (z) | =∞.

Theorem 7.18,  Page 242.  The function  f (z)  has an essential singularity at  z = α  if and only if   (lim_(z -> α))| 
        f (z) |   does not exist.

Example 7.15,  Page 240.  Find the first few terms of the Maclaurin series 
for   f(z) = sec(z)   and use it to compute  (f^((4)))(0).

\(Clear[f, S, z]; 
f[z_]\  = \ Sec[z]; 

    S[z_]\  = \ Normal[Series[Sec[z], \ {z, 0, 12}]]; 

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

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

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

    Print["f''''[0]] = ", \(\(\(\(f'\)'\)'\)'\)[0]]; 

    Print["S''''[0]] = ", \(\(\(\(S'\)'\)'\)'\)[0]]; \)

Example 7.16,  Page 242.  Show that the function   g(z) = (e^(((-1)/(z^2)))) is not continuous at   z = 0.

\(Clear[g, lim1, lim2, x, y, z]; 

    g[z_]\  = \ E\^\(\(-1\)\/z\^2\); 

    lim1\  = \ Limit[g[x\  + \ 0\ I], \ x -> 0]; 

    lim2\  = \ Limit[g[0\  + \ I\ y], \ y -> 0]; 

    val\  = \ MapAll[Together, 
        ComplexExpand[g[x + I\ y], TargetFunctions -> {Re, Im}]]; 

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

    Print["   g[x+Iy] = ", g[x + I\ y]]; 

    Print["   g[x+Iy] = ", val]; 

    Print[\*""\!\(lim\+\(x -> 0\)\) g[x+0I] = "", lim1]; 

    Print[\*""\!\(lim\+\(y -> 0\)\) g[0+Iy] = "", lim2]; \)

Since the limits are not the same,  g(z) = (e^(((-1)/(z^2))))  is not continuous at   z = 0.

GoTo Chapter 7