COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

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COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
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 CHAPTER 4 SEQUENCES, JULIA and MANDELBROT SETS, and Power Series

Section 4.1 Sequences and Series

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. For convenience, we at times use the term sequence rather than c omplex sequence .

 

If we wish a function s to represent an arbitrary sequence, we could specify it by writing s(1) = z[1] , s(2) = z[2] , and so on. The values z[1] , z[2] , z[3] , ..., are called the terms of a sequence, and mathematicians, being generally lazy when it comes to things like this, often refer to z[1] , z[2] , z[3] , etc., as the sequence itself, even though they are really speaking of the range of the sequence when they do this. Mathematicians are also not so fussy about starting a sequence at z[1] , so that z[0] , z[1] , z[2] , ..., etc., would also be acceptable notation, provided all terms were defined.

 

Definition 4.1: Limit of a sequence

 

The expression

 

Limit(z[n],n = infinity) = zeta means that

 

for any real number epsilon >0, there corresponds a positive integer N[epsilon] which depends on epsilon such that z[n] epsilon D[epsilon] ( zeta ) whenever n > N[epsilon] .

 

Theorem 4.1 Let z[n] = x[n]+i*y[n] and zeta = u+i*v . Then,

 

Limit(z[n],n = infinity) = zeta if and only if Limit(x[n],n = infinity) = u and Limit(y[n],n = infinity) = v .

 

Example 4.1, Page 128. Find limit((sqrt(n)+i*(n+1))/n,n = infinity) .

> L:='L': n:='n': Xn:='Xn': Yn:='Yn': Zn:='Zn': z:='z':
Xn := sqrt(n)/n:
Yn := (n+1)/n:
Zn := Xn + I*Yn:
L := limit(Zn, n=infinity):
z[n] = Zn; ` `;
limit( z[n], n=infinity) = L;

z[n] = 1/(sqrt(n))+I*(n+1)/n

` `

limit(z[n],n = infinity) = I

 

 

Example 4.2, Page 128. Show that the sequence z[n] = (1+i)^n diverges.

> n:='n': L:='L': n:='n': Zn:='Zn': z:='z':
Zn := (1 + I)^n:
L := limit(Zn, n=infinity):
z[n] = Zn; ` `;
limit( z[n], n=infinity) = L;

z[n] = (1+I)^n

` `

limit(z[n],n = infinity) = limit((1+I)^n,n = infini...

Maple did NOT find the limit!

Now look at the real and imaginary parts of z[n] .

> L:='L': n:='n': Xn:='Xn': Yn:='Yn': Zn:='Zn':
x:='x': y:='y': z:='z':
Xn := sqrt(2)^n * cos(n*pi/4):
Yn := sqrt(2)^n * sin(n*pi/4):
Zn := Xn + I*Yn:
L := limit(Xn, n=infinity) + I*limit(Yn, n=infinity):
`The general term is:`;
z[n], ` = `,x[n] + y[n] = Zn; ` `;
limit( z[n], n=infinity) = L;

`The general term is:`

z[n], ` = `, x[n]+y[n] = (sqrt(2))^n*cos(1/4*n*pi)+...

` `

limit(z[n],n = infinity) = undefined+undefined*I

 

 

Theorem 4.2 If {z[n]} is a Cauchy sequence, then {z[n]} converges.

 

 

Definition 4.3: Infinite Series

The formal expression Sum(z[k],k = 1 .. infinity) is called an infinite series , and z[1] , z[2] , etc., are called the terms of the series.

 

If there is a complex number S for which

 

S = Limit(S[n],n = infinity) = Limit(Sum(z[k],k = 1 .. n),n = infinity) ,

 

we will say that the infinite series Sum(z[k],k = 1 .. infinity) converges to S , and that S is the sum of the infinite series.

 

When this happens, we write S = Sum(z[k],k = 1 .. infinity) .

 

 

Theorem 4.3 Let z[n] = x[n]+i*y[n] and S = U+i*V . Then

 

S = Sum(z[n],n = 1 .. infinity) = Sum(x[n]+i*y[n],n = 1 .. infinity)

 

if and only if U = Sum(x[n],n = 1 .. infinity) and V = Sum(y[n],n = 1 .. infinity) .

 

Theorem 4.4 (n-th term test)

 

If Sum(z[n],n = 1 .. infinity) is a convergent complex series, then Limit(z[n],n = infinity) = 0 .

 

Example 4.3, Page 130. Show that the series sum((1+i*n*(-1)^n)/(n^2),n = 1 .. infinity) converges.

> n:='n': S:='S': Xn:='Xn': Yn:='Yn': Zn:='Zn': z:='z':
Xn := 1/n^2:
Yn := (-1)^n/n:
Zn := Xn + I*Yn:
z[n] = Zn;
S := sum(Xn, n=1..infinity) + I*sum(Yn, n=1..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) =
Sum(Xn, n=1..infinity)+ I*Sum(Yn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;

z[n] = 1/(n^2)+I*(-1)^n/n

`Find the sum of the series:`

Sum(z[n],n = 1 .. infinity) = Sum(1/(n^2),n = 1 .. ...

Sum(z[n],n = 1 .. infinity) = 1/6*Pi^2-I*ln(2)

Example 4.4, Page 130. Show that the series sum(((-1)^n+i)/n,n = 1 .. infinity) diverges.

> n:='n': S:='S': Xn:='Xn': Yn:='Yn': Zn:='Zn': z:='z':
Xn := (-1)^n /n:
Yn := 1/n:
Zn := Xn + I*Yn:
z[n] = Zn;
S := sum(Xn, n=1..infinity) + I*sum(Yn, n=1..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) =
Sum(Xn, n=1..infinity)+ I*Sum(Yn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;

z[n] = (-1)^n/n+I/n

`Find the sum of the series:`

Sum(z[n],n = 1 .. infinity) = Sum((-1)^n/n,n = 1 .....

Sum(z[n],n = 1 .. infinity) = -ln(2)+infinity*I

 

Example 4.5, Page 131. Show that the series sum((1+i)^n,n = 1 .. infinity) diverges.

> n:='n': S:='S': Zn:='Zn': z:='z':
Zn := (1 + i)^n:
z[n] = Zn;
S := sum(Zn, n=1..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) = Sum(Zn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;
`But this is NOT the right answer !`;

z[n] = (1+i)^n

`Find the sum of the series:`

Sum(z[n],n = 1 .. infinity) = Sum((1+i)^n,n = 1 .. ...

Sum(z[n],n = 1 .. infinity) = -(1+i)/i

`But this is NOT the right answer !`

 

BAD news! MAPLE substituted into a divergent geometric series ! It substituted Z = 1+i into the
formula 1/(1-Z)-1 . You should always check to see if 1 < abs(Z) , or risk getting a wrong answer!

 

> n:='n': S:='S': Z:='Z': Zn:='Zn': z:='z':
Zn := (1 + I)^n:
Z := 1 + I:
S := 1/(1-Z) - 1:
`1/(1-Z) - 1 ` = S;
`where Z ` = Z;
`The general term is:`;
z[n] = Zn; ` `;
`But`;
1 < `|1+I| `;
1 < abs(Z);
1 < evalf(abs(Z));
`Therefore the series diverges.`;

`1/(1-Z) - 1 ` = -1+I

`where Z ` = 1+I

`The general term is:`

z[n] = (1+I)^n

` `

But

1 < `|1+I| `

1 < sqrt(2)

1 < 1.414213562

`Therefore the series diverges.`

 

End of Section 4.1.