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 4  SEQUENCES, JULIA and MANDELBROT SETS, and POWER SERIES

c = q = S = t = x = y = z = 
                    0; 
Clear[a, absZn, c, cx, cy, dot, i, juliaC, k, lim, limX, limY, limZ, 
        list, mandelbrotC, n, newton, Pn, q, Qn, R, s, S, Sn, set, sum, sumN, 
        sumX, sumY, sumZ, t, table, x, Xn, y, Yn, z, Z, 
        Zn]; 
Needs["Graphics`ComplexMap`"]; 
Needs["Graphics`Colors`"];

Section 4.1  Sequences and Series

In this chapter we learn the basics for complex sequences and series.  We also explore an application of these ideas in what has popularly ∞come to be known as chaotic processes.

	In formal terms, a complex sequence is a function whose domain is the positive integers and whose range is a subset of the complex numbers.  The following are examples of sequences: 

	f (n) = (2 - 
              (1/n)) + i (5 + 
                (1/n))      (n = 1, 2, 3,  ... ) 

	g (n) = (2 - 
              (1/n)) + i (5 + 
                (1/n))      (n = 1, 2, 3,  ... ) 

	\(h \((n)\) = 
        5 + 3  i + \(\(\((1\/\(1 + i\))\)\^n\)\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \)\((n = 1, 2, 3,  ... )\)\(\ \)\)\)

	\(r \((n)\) = 
        5 + 3  i + \(\(\((1\/4 + i\/2)\)\^n\)\(\ \ \ \ \ \ \ \ \ \)\((n = 1, 2, 3,  ... )\)\(\ \)\)\)
	
	For convenience, we at times use the term sequence rather than complex sequence.  If we wish a function  s  to represent an arbitrary sequence, we could specify it by writing  s (1) = (Subscript[z, 1]), s (2) = (Subscript[z, 2]), s (3) = (Subscript[z, 3]),  and so on.  The values   (Subscript[z, 1]), (Subscript[z, 2]), (Subscript[z, 3]),  ... ,  are called the terms of a sequence, and mathematicians, being generally lazy when it comes to things like this, often refer to (Subscript[z, 1]), (Subscript[z, 2]), (Subscript[z, 3]), etc. as the sequence itself, even though they are really speaking of the range of the sequence when they do this.  You will usually see a sequence written as  {(Subscript[z, n])}_(n = 1)^∞, {(Subscript[z, n])}_1^∞, or when the indices are understood, as {(Subscript[z, n])}.  Mathematicians are also not so fussy about starting a sequence at  (Subscript[z, 1]), so that  {(Subscript[z, n])}_(n = -1)^∞, {(Subscript[z, n])}_(n = 0)^∞, etc., would also be acceptable notation, (Subscript[n, □])provided all terms were defined. 

	Informally, the sequence  {(Subscript[z, n])}_1^∞  has  ζ  as its limit as  n  approaches infinity, provided the terms  (Subscript[z, n])  can be made as close as we please to  ζ  by making  n  large enough.  When this happens, we write 

	(lim_(n -> ∞)) (Subscript[z, n])=ζ    or    (Subscript[z, n])->ζ   as  n  -> ∞

If  (lim_(n -> ∞)) (Subscript[z, n]) = ζ,  we say that the sequence  {(Subscript[z, n])}_1^∞  converges to  ζ.  

	However, we need a rigorous definition, if we are to do honest mathematics.  Thus, we have the following.

Definition 4.1 (Limit of a Sequence), Page 126.  (lim_(n -> ∞)) (Subscript[z, n]) = ζ  means that for any real number  ϵ > 0  there corresponds a positive integer  (Subscript[N, ϵ])  (which depends on  ϵ)  such that  (Subscript[z, n]) ∈ (Subscript[D, ϵ]) (ζ)  whenever  n > (Subscript[N, ϵ]),  in other words  |ζ - (Subscript[z, n])|<∈  whenever  n > (Subscript[N, ϵ]).

Remark 4.1. The reason we use the notation (Subscript[N, ϵ]) is to emphasize the fact that this number depends on our choice of ϵ. Sometimes it will be convenient to drop the subscript.

In form, Definition 4.1 is exactly the same as the corresponding definition for limits of real sequences. In fact, there is a simple criterion that casts the convergence of complex sequences in terms of the convergence of real sequences.

Theorem 4.1,  Page 126.  Let  (Subscript[z, n]) = (Subscript[x, n]) + i (Subscript[y, n])  and  ζ = u + i v.  Then 

		(lim_(n -> ∞)) (Subscript[z, n]) = ζ  

	if and only if  

		(lim_(n -> ∞)) (Subscript[x, n])=u  and  (lim_(n -> ∞)) (Subscript[y, n])=v.

Proof of Theorem 4.1,  see text Page 126.

Example 4.1,  Page 128.  Find the limit of the sequence  {(Subscript[z, n])} = {(((Sqrt[n])  +  i ( n + 1))/n)}.

Solution 4.1.

Enter the formula for the terms of the sequence.

Remove[x, y, z]; 
Clear[limX, limY, limZ, n]; 
(Subscript[x, n])_  =  ((Sqrt[n])/n); 
(Subscript[y, n])_  =  ((n + 1)/n); 
(Subscript[z, n])_  =  
        (Subscript[x, n])  +  I (Subscript[y, n]); 
table = Table[(Subscript[z, n]), {n, 1, 9}]; 
table = Append[table, "..."]; 
limX  =  Limit[(Subscript[x, n]),  n -> ∞]; 
limY  =  Limit[(Subscript[y, n]),  n -> ∞]; 
limZ  =  
        Limit[(Subscript[z, n]),  n -> ∞]; 
\(\(Print[\*""\!\(z\_n\) = "", 
        z\_n];\)\ \)
\(\(Print[\*""\!\(x\_n\) = "", 
        x\_n];\)\ \)
\(\(Print[\*""\!\(y\_n\) = "", 
        y\_n];\)\ \)
Print[""]; 
\(\(Print[\*""(lim_(n -> ∞)) (Subscript[x, n]) = "", \*""(lim_(n -> ∞)) "", x\_n, " = ", 
        limX];\)\ \)
\(\(Print[\*""(lim_(n -> ∞)) (Subscript[y, n]) = "", \*""(lim_(n -> ∞)) "", y\_n, " = ", 
        limY];\)\ \)
Print[""]; 
\(\(Print[\*""{\!\(z\_n\)} = "", 
        table];\)\ \)
Print[""]; 
\(\(Print[\*""{\!\(z\_n\)} = "", 
        N[table]];\)\ \)
Print[""]; 
\(\(Print[\*""(lim_(n -> ∞)) (Subscript[z, n]) = "", \*""(lim_(n -> ∞)) ("", 
        z\_n, ")"];\)\ \)
\(\(Print[\*""\!\(lim\+\(n -> ∞\)\) \!\(z\_n\) = "", 
        limX, " + I (", limY, ") = ", 
        limZ];\)\ \)

Use Mathematica to plot some of the terms in the sequence.

set  =   Table[{N[(Subscript[x, n])], N[(Subscript[y, n])]}, {n, 1, 
            50}]; 
dot = 
        Graphics[{PointSize[0.025], Magenta, 
            Point[{0, 1}]}]; 
\(\(gr = 
        ListPlot[set, Prolog -> {PointSize[0.02], Green}, 
          PlotRange -> {{0.0, 1.0}, {0.0, 2.0}}, 
          AxesLabel -> {"x", "y"}, 
Ticks -> {Range[0, 1, 1\/2], 
              Range[0, 2, 1\/2]}];\)\ \)
Show[gr, dot]; 
\(\(Print[\*""(lim_(n -> ∞)) (Subscript[z, n]) = (lim_(n -> ∞)) ("", z\_n, ") = ", 
        limX, " + I (", limY, ") = ", 
        limZ];\)\ \)

We see that the limit of the sequence  \({z\_n} = {\(\@n\  + \ i \((\ n + 1)\)\)\/n}\ \ \ is\ \ \ i\).  However, the real part is converging slowly to 0  and the imaginary part is converging a little faster to  i.

Example 4.2,  Page 128.  Show that the sequence {(((1 + i))^n)} diverges.

Solution 4.2.

Enter the formula for the terms of the sequence, and determine if the sequence converges or diverges.

Remove[x, y, z]; 
Clear[limX, limY, limZ, n]; 
(Subscript[z, n])_ =  (((1  +  I))^n); 
table = Table[(Subscript[z, n]), {n, 1, 10}]; 
table = Append[table, "..."]; 
lim  =  
        Limit[(Subscript[z, n]),  n -> ∞]; 
\(\(Print[\*""{\!\(z\_n\)} = "", 
        table];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(lim\+\(n -> ∞\)\) "", 
        z\_n, " = ", lim];\)\ \)

Investigate the limit in more detail.

(Subscript[x, n])_ = 
        ComplexExpand[Re[(Subscript[z, n])]]; 
(Subscript[y, n])_ = ComplexExpand[Im[(Subscript[z, n])]]; 
limX  =  Limit[(Subscript[x, n]),  n -> ∞]; 
limY  =  Limit[(Subscript[y, n]),  n -> ∞]; 
limZ  =  limX  +  I limY; 
\(\(Print[\*""(lim_(n -> ∞)) (Subscript[x, n]) = "", \*""(lim_(n -> ∞)) "", x\_n, " = ", 
        limX];\)\ \)
\(\(Print[\*""(lim_(n -> ∞)) (Subscript[y, n]) = "", \*""(lim_(n -> ∞)) "", y\_n, " = ", 
        limY];\)\ \)
Print[""]; 
\(\(Print[\*""(lim_(n -> ∞)) (Subscript[z, n]) = "", \*""(lim_(n -> ∞)) "", 
        z\_n];\)\ \)
Print["       = ", limX, " + \[ImaginaryI] (", 
        limY, ")"]; 
\(\(Print[\*""(lim_(n -> ∞)) (Subscript[z, n]) = "", \*""(lim_(n -> ∞)) "", 
        z\_n];\)\ \)
Print["       = ", limZ];

Therefore, the limit of the sequence does NOT exist, and exhibits "divergent oscillations."

Use Mathematica to plot some of the terms in the sequence.

set  =   Table[{N[(Subscript[x, n])], N[(Subscript[y, n])]}, {n, 1, 
            7}]; 
\(\(ListPlot[set, Prolog -> {PointSize[0.02], Blue}, 
        PlotRange -> {{\(-4.3\), 8.0}, {\(-9.0\), 2.5}}, 
        AxesLabel -> {"x", "y"}, 
AspectRatio -> 1, 
        Ticks -> {Range[\(-4\), 8, 2], 
            Range[\(-8\), 2, 2]}];\)\ \)
\(\(Print[\*""{\!\(z\_n\)} = "", 
        table];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(lim\+\(n -> ∞\)\) "", 
        z\_n, " = ", limX, " + I ", 
        limY];\)\ \)

We see that the sequence  {(Subscript[z, n])} = {(((1 + i))^n)}  is divergent.

As with real numbers, we also have the following definition.

Definition 4.2, Page 128.  The sequence  {(Subscript[z, n])}  is said to be a Cauchy Sequence if for every  ϵ > 0  there exists a positive integer  (Subscript[N, ϵ]),  such that if  m, n > (Subscript[N, ϵ]),  then  |(Subscript[z, m]) - (Subscript[z, n])|<∈,  or, equivalently,  (Subscript[z, m]) - (Subscript[z, n])  belongs to the disk  (Subscript[D, ϵ]) (0).

The following should now come as no surprise.

Theorem 4.2,  (Cauchy Sequences Converge)Page 128.  If  {(Subscript[z, n])}  is a Cauchy sequence, then {(Subscript[z, n])}  converges.

Proof of Theorem 4.2,  see text Page 128.

If we have a complex sequence {(Subscript[z, n])}, we can form a new sequence {(Subscript[S, n])}, called the sequence of partial sums, in the following way:  

				(Subscript[S, 1]) = (Subscript[z, 1]),  
				(Subscript[S, 2]) = (Subscript[z, 1]) + (Subscript[z, 2]),  
				(Subscript[S, 3]) = (Subscript[z, 1]) + (Subscript[z, 2]) + (Subscript[z, 3]), 
				     ⋮  
				(Subscript[S, n]) = (Subscript[z, 1]) + (Subscript[z, 2]) +  ...  + 
            (Subscript[z, n]) = (∑_(k = 1)^∞) (Subscript[z, k]),
				     ⋮

Definition 4.3 (Infinite Series), Page 129.  The formal expression (∑_(k = 1)^∞) (Subscript[z, k]) = (Subscript[z, 1]) + (Subscript[z, 2]) +  ...  + 
            (Subscript[z, n]) +  ...   is called an infinite series, and (Subscript[z, 1]), (Subscript[z, 2]),  ... , are called the terms of the series.

If there is a complex number S for which 

			S = (lim_(n -> ∞)) (Subscript[S, n]) = 
          (lim_(n -> ∞))(∑_(k = 1)^n) (Subscript[z, k]), 
			
we will say that the infinite series (∑_(k = 1)^∞) (Subscript[z, k]) converges to S, and that S is the sum of the infinite series. When this happens, we write 

			S = (∑_(k = 1)^∞) (Subscript[z, k]).  

	The series (∑_(k = 1)^∞) (Subscript[z, k]) is said to be absolutely convergent provided that the (real) series of magnitudes (∑_(k = 1)^∞)|(Subscript[z, k])| converges.  

	If a series does not converge, we say that it diverges.

Remark 4.2. It is important to note that the first finitely many terms of a series do not affect its convergence or divergence and that in this respect the beginning index of a series is irrelevant. Thus, we will without comment conclude that if a series (∑_(k = N + 1)^∞) (Subscript[z, k]) converges, then so does (∑_(k = 0)^∞) (Subscript[z, k]), where (Subscript[z, 0]), (Subscript[z, 1]), (Subscript[z, 2]),  ... , (Subscript[z, N]) is any finite collection of terms. A similar remark holds for determining divergence of a series.

As one might expect, many of the results concerning real series carry over to the complex case. We now give several of the more standard theorems along with examples of how they are used.

Theorem 4.3,  Page 130.  Let  (Subscript[z, n]) = (Subscript[x, n]) + i (Subscript[y, n])  and  S = U + i V.  Then  

			\(S = \(∑\+\(n = 1\)\%∞ z\_n = ∑\+\(n = 1\)\%∞\((x\_n + i\ y\_n)\)\)\)  converges 

if and only if both  
			\(U = \(∑\+\(n = 1\)\%∞ x\_n\ \ and\ \ V = ∑\+\(n = 1\)\%∞ y\_n\)\)  converge.

Proof of Theorem 4.3,  see text Page 130.

Theorem 4.4,  Page 130.  If  (∑_(n = 1)^∞) (Subscript[z, n])  is a convergent complex series, then  (lim_(n -> ∞)) (Subscript[z, n]) = 0.

Proof of Theorem 4.4, Page 130. The proof is left as an exercise.

Example 4.3,  Page 130.  Show that the series  \(∑\+\(n = 1\)\%∞\( 1 + i\ n \((\(-1\))\)\^n\)\/n\^2\)  is convergent.

Solution 4.3.

Enter the formula for the series, and determine if the series converges or diverges.

Clear[k, n, sumX, sumY, sumZ]; 
sumX =(∑_(n = 1)^∞) (1/(n^2)); 
sumY = (∑_(n = 1)^∞)(( n (((-1))^n))/(n^2)); 
sumZ  =  sumX  +  I sumY; 
Print["         (∑_(n = 1)^∞)(1/(n^2)) = ",sumX]; 
Print["    (∑_(n = 1)^∞)((n (((-1))^n))/(n^2)) = ",sumY]; 
Print[""]; 
\(Print[\*""(∑_(n = 1)^∞)((1 + In (((-1))^n))/(n^2)) = "", sumZ];\) 
\(Print[\*""(∑_(n = 1)^∞)((1 + In (((-1))^n))/(n^2)) = "", N[sumZ]];\)

Use Mathematica to construct some of the partial sums of the infinite series.

(Subscript[X, n])_  = (∑_(k = 1)^n) (1/(k^2)); 
(Subscript[Y, n])_  = (∑_(k = 1)^n)((((-1))^k)/(k^2)); 
(Subscript[S, n])_ = (∑_(k = 1)^n) (1/(k^2)) + I  (∑_(k = 1)^n)((((-1))^k)/(k^2)); 
table = Table[FullSimplify[(Subscript[S, n])], {n, 1, 10}]; 
table = Append[table, "..."]; 
\(Print[\*""{\!\(S\_n\)} = "", table];\) 
Print[""]; 
\(Print[\*""{\!\(S\_n\)} = "", N[table]];\) 
Print[""]; 
\(Print[\*""(Subscript[S, n]) = ((∑_(k = 1)^n)(1/(k^2)))+I((∑_(k = 1)^n)((((-1))^k)/(k^2))) = "", S\_n];\) 
Print[""]; 
\(Print[\*"" (∑_(k = 1)^n)(1/(k^2))+I(∑_(k = 1)^n)((((-1))^k)/(k^2)) = "", 
          sumZ];\)

Use Mathematica to plot some of the partial sums of the infinite series.

set  =   Table[{(Subscript[X, n]), (Subscript[Y, n])}, {n, 1, 
            40}]; 
\(\(ListPlot[set, Prolog -> {PointSize[0.02], Magenta}, 
        PlotRange -> {{0.0, 1.7}, {\(-1.05\), 
              0.05}}, 
AxesLabel -> {"x", "y"}, 
Ticks -> {Range[0, 2, 0.2], 
            Range[\(-1\), 0, 0.2]}];\)\ \)
\(\(Print[\*""\!\(lim\+\(n -> ∞\)\) \!\(S\_n\) = "", 
        sumZ];\)\ \)
\(\(Print[\*""\!\(lim\+\(n -> ∞\)\) \!\(S\_n\) = "", 
        N[sumZ]];\)\ \)
Print[""]; 
\(\(Print[\*""(∑_(k = 1)^n)((1 + I (((-1))^k))/(k^2)) = "", sumZ];\)\ \)
Null

Therefore, we see that the infinite series converges, indeed  \(∑\+\(n = 1\)\%∞\( 1\  + \ i\ n \((\(-1\))\)\^n\)\/n\^2 = π\^2\/6 - i\ Log[2]\).

Example 4.4,  Page 130.  Show that the series  \(∑\+\(n = 1\)\%∞\(\((\(-1\))\)\^n + i\)\/n\)  is divergent.

Solution 4.4.

Enter the formula for the series, and determine if the series converges or diverges.

Clear[n, sumN, sumX, sumY, sumZ]; 
sumX  =  (∑_(n = 1)^∞)((((-1))^n)/n); 
sumY  =  (∑_(n = 1)^∞) (1/n); 
sumZ  =  sumX  +  I sumY; 
\(\(Print[\*""(∑_(n = 1)^∞)((((-1))^n)/n) = "", sumX];\)\ \)
\(\(Print[\*""\!\(∑\+\(n = 1\)\%∞\)\!\(1\/n\) = "", 
        sumY];\)\ \)
Print[""]; 
\(\(Print[\*""(∑_(n = 1)^∞)(((((-1))^n) + I)/n) = "", 
        sumZ];\)\ \)

Remark.  Since the answer returned for the imaginary part was  (∑_(n = 1)^∞) (1/n),  this means that a sum was not found.
It is known that the partial sums of the harmonic series grow slowly without bound.  
For example, adding 10,  100,  1000  and  10000 terms yields:

\(
\(Print[\*""(∑_(n = 1)^m)(1/n) = "", ∑\+\(n = 1\)\%m 1\/n];\)\ \)
Print[""]; 
\(\(For[i = 1, 
        i ≤ 4, \(i++\), 
\ \ Print["m = ", 
          10\^i, \*"",     \!\(∑\+\(n = 1\)\%m\)\!\(1\/n\) = "", 
          N[HarmonicNumber[10\^i]]]];\)\ \)

Therefore, the series slowly diverges to ∞.

Example 4.5,  Page 131.  Show that the series (∑_(n = 1)^∞)(((1 + i))^n) is divergent.

Solution 4.5.

Enter the formula (Subscript[z, n]) = (((1 + i))^n) for the terms of the sequence, and look at the modulus.

Remove[x, y, z]; 
Clear[limX, limY, limZ, n]; 
(Subscript[z, n])_ =  (((1  +  I))^n); 
(Subscript[m, n])_ = Abs[(Subscript[z, n])]; 
(Subscript[M, n])_ = ComplexExpand[(Subscript[m, n])]; 
lim  =  
        Limit[(Subscript[m, n]),  n -> ∞]; 
Print[""]; 
\(\(Print[\*""\!\(lim\+\(n -> ∞\)\) |"", 
        z\_n, \*""| = \!\(lim\+\(n -> ∞\)\) "", 
        m\_n];\)\ \)
\(\(Print[\*""\!\(lim\+\(n -> ∞\)\) |"", 
        z\_n, \*""| = \!\(lim\+\(n -> ∞\)\) "", 
        M\_n];\)\ \)
\(\(Print[\*""\!\(lim\+\(n -> ∞\)\) |"", 
        z\_n, "| = ", lim];\)\ \)

Hence (lim_(n  ->  ∞)) (Subscript[z, n]) ≠ 0, and the series is divergent.

Aside.  Mathematica will indicate that there is a problem with convergence.

sum = (∑_(n = 1)^∞)(((1 + I))^n);   
\(\(Print[\*""(∑_(n = 1)^∞)(((1 + I))^n) = "", sum];\)\ \)

Which is comforting to see that Mathematica knows the answer too!

Theorem 4.5,  Page 131.  Let  \(∑\+\(n = 1\)\%∞ z\_n\ \ and\ \ \(∑\+\(n = 1\)\%∞ w\_n\)\)  be convergent series, and let  c  be a complex number.  Then  

			(∑_(n = 1)^∞) c (Subscript[z, n]) = 
        c (∑_(n = 1)^∞) (Subscript[z, n])  
		and  
			\(∑\+\(n = 1\)\%∞\((z\_n + 
              w\_n)\) = ∑\+\(n = 1\)\%∞ z\_n + ∑\+\(n = 1\)\%∞ w\_n\).

Proof of Theorem 4.5,  see text Page 131. The proof of this theorem is left as an exercise.

Definition 4.4, Page 131.  Let  (∑_(n = 1)^∞) (Subscript[a, n])  and  (∑_(n = 1)^∞) (Subscript[b, n])  be convergent series, where (Subscript[a, n])  and  (Subscript[b, n])  are complex numbers.  The Cauchy Product of the two series is defined to be the series  (∑_(n = 1)^∞) (Subscript[c, n]),  where  (Subscript[c, n]) = (∑_(k = 1)^n) (Subscript[a, k]) (Subscript[b, n - k]).

Theorem 4.6,  Page 131.  If the Cauchy product converges, then  

			\(∑\+\(n = 1\)\%∞ c\_n = \((∑\+\(n = 1\)\%∞ a\_n)\) \((∑\+\(n = 1\)\%∞ b\_n)\)\),  
		where  
			(Subscript[c, n]) = (∑_(k = 1)^n) (Subscript[a, k]) (Subscript[b, n - k]).

Proof of Theorem 4.6,  see text Page 131.  The proof can be found in a number of texts, for example, Infinite Sequences and Series, by Konrad Knopp (translated by Frederick Bagemihl;  New York: Dover, 1956).

Theorem 4.7,  Page 132.  (Comparison Test)  Let  (∑_(n = 1)^∞) (Subscript[M, n])  be a convergent series of real nonnegative terms.  If  {(Subscript[z, n])}  is a sequence of complex numbers and  |(Subscript[z, n])|≤ (Subscript[M, n])  holds for all  n, then  (∑_(n = 1)^∞) (Subscript[z, n])  converges.

Proof of Theorem 4.7,  see text Page 132.

Corollary 4.1, Page 132.  If  (∑_(n = 1)^∞)|(Subscript[z, n])|, then  (∑_(n = 1)^∞) (Subscript[z, n])  converges.  In other words, absolute convergence implies convergence for complex series as well as for real series.

Proof of Corollary 4.1, see text Page 132.

Example 4.6,  Page 132.  Show that the series  \(∑\+\(n = 1\)\%∞\((3\  + \ 4  i)\)\^n\/\(5\^n\ n\^2\)\)  is convergent.

Solution 4.6.

Enter the formula for the series, and determine if the series converges or diverges.

Remove[x, y, z]; 
Clear[absZn, n, sum, sumX, sumY, sumZ]; 
(Subscript[z, n])_  =  ((((3 + 4  I))^n)/((5^n) (n^2))); 
(Subscript[x, n])_  =  ComplexExpand[Re[(Subscript[z, n])]]; 
(Subscript[y, n])_  =  ComplexExpand[Im[(Subscript[z, n])]]; 
sumX  =  (∑_(n = 1)^∞) (Subscript[x, n]); 
sumY  =  (∑_(n = 1)^∞) (Subscript[y, n]); 
sumZ  =  (∑_(n = 1)^∞) (Subscript[z, n]); 
\(\(Print[\*""\!\(z\_n\) = "", z\_n];\)\ \)
\(\(Print[\*""\!\(z\_n\) = "", ComplexExpand[z\_n]];\)\ \)
\(\(Print[\*""\!\(x\_n\) = "", x\_n];\)\ \)
\(\(Print[\*""\!\(y\_n\) = "", 
        y\_n];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(∑\+\(n = 1\)\%∞\)\!\(x\_n\) = "", 
        sumX];\)\ \)
Print["     = ", Chop[N[sumX]]]; 
\(\(Print[\*""\!\(∑\+\(n = 1\)\%∞\)\!\(y\_n\) = "", 
        sumY];\)\ \)
Print["     = ", Chop[N[sumY]]]; 
Print[""]; 
\(\(Print[\*""(∑_(n = 1)^∞)((((3 + 4  I))^n)/((5^n)*(n^2))) = "", 
        sumZ];\)\ \)
\(\(Print[\*""(∑_(n = 1)^∞)((((3 + 4  I))^n)/((5^n)*(n^2))) = "", 
        N[sumZ]];\)\ \)

Hence we see that the series converges and that  \(∑\+\(n = 1\)\%∞\((3\  + \ 4  i)\)\^n\/\(5\^n\ n\^2\) = 
        PolyLog[2, 3\/5 + \(4\ i\)\/5]\).  

Or we could use Corollary 4.1 and show that the series converges absolutely.

absZn  =  
        ComplexExpand[Abs[(Subscript[z, n])]]; 
sum   =   (∑_(n = 1)^∞) (1/(n^2)); 
Print["|", (Subscript[z, n]), "| = ", 
        absZn]; 
Print[""]; 
\(\(Print[\*""(∑_(n = 1)^∞)|((((3 + 4  I))^n)/((5^n)*(n^2)))| = "", 
        sum];\)\ \)

The series of absolute values converges, therefore the series converges.

Use Mathematica to construct some of the partial sums of the infinite series.

(Subscript[S, n])_  =  (∑_(k = 1)^n)((((3 + 4  I))^k)/((5^k) (k^2))); 
table = 
        Table[FullSimplify[(Subscript[S, n])], {n, 1, 10}]; 
table = Append[table, "..."]; 
\(\(Print[\*""{\!\(S\_n\)} = "", 
        table];\)\ \)
Print[""]; 
\(\(Print[\*""{\!\(S\_n\)} = "", 
        N[table]];\)\ \)
Print[""]; 
\(\(Print[\*""(Subscript[S, n]) = (∑_(k = 1)^n)((((3 + 4  I))^k)/((5^k)*(k^2))) = "", 
        S\_n];\)\ \)

Use Mathematica to plot some of the partial sums of the infinite series.

set  =   Table[{Re[(Subscript[S, n])], Im[(Subscript[S, n])]}, {n, 1, 
            10}]; 
ListPlot[set, 
        Prolog -> {PointSize[0.03], 
            Green}, 
PlotRange -> {{0, 0.7}, {0, 
              1.1}}, 
AxesLabel -> {"x", "y"}, 
        AspectRatio -> 
          (1.1/0.7), 
Ticks -> {Range[0, 1, 0.2], 
            Range[0, 1, 0.2]}]; 
\(\(Print[\*""{\!\(S\_n\)} = "", 
        N[table]];\)\ \)
Print[""]; 
\(\(Print[\*""(∑_(n = 1)^∞)((((3 + 4  I))^n)/((5^n)*(n^2))) = "", sumZ, " = ", 
        N[sumZ]];\)\ \)

Therefore, we see that the series converges, and we have  \(\(\(\ \)\(∑\+\(n = 1\)\%∞\((3\  + \ 4  i)\)\^n\/\(5\^n\ n\^2\) = 
        PolyLog[2, 3\/5 + \(4\ i\)\/5]\)\(\ \)\)\).

Library Research Experience for Undergraduates

Project I. Write a report on series.
	
1. Agnew, R. P. and R. P. Boas, ''An Integral Test for Convergence (in Math. Notes),'' Am. Math. M., Vol. 56, No. 10. (Dec., 1949), pp. 677-678, Jstor.   

2. Barry, P. D. and D. J. Hurley, (1991), ''On Series of Partial Fractions,'' Am. Math. M., V. 98, No. 3, pp. 240-242.

3. Gluchoff, Alan D., (1993), ''Complex Power Series -- a Vector Field Visualization,'' Math. Mag., V. 66, No. 3, pp. 189-191.

4. Lee, Sanboh, and J. C. M. Li, (1988), ''Summation of Infinite Series Related to Roots of Certain Functions,'' Int. J. of Math. Ed. in Sci. and Tech., V. 19, No. 1, pp. 89-93.

5. Marafino, John and Timothy J. McDevitt, (1995) ''Convergence of Complex Continued Fractions,'' Math. Mag., Vol.68, No 3, pp.202-208.

Section 4.1  Exercises for Sequences and Series
See textbook page 132.

Section 4.2	Julia and Mandelbrot Sets

Section 4.3	Geometric Series and Convergence Theorems

Section 4.4	Power Series Functions

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