1. No category

c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
1
Exercises involving Taylor’s series and Laurent series
1. Find the circle of convergence of the following power series justifying your answer
in each case.
(a)
∞
X
(z − i)n
2n
n=0
.
(b)
∞
X
(z + 4i)2n (n − 1)2 .
n=0
2. Suppose that a pseudo random number generator generates a sequence a0 , a1 , · · ·
with |an | ≤ 1. Find the circle of convergence of the power series
∞
X
(2 + an )z n .
n=0
3. Obtain the Maclaurin series for z 3 sin(2z) and state where it converges.
4. Let
∞
X
an z n
n=0
be the Maclaurin series of f (z) analytic at the origin. Show that
∞
X
an z 2n
0
is the Maclaurin series of g(z) = f (z 2 ).
5. Prove that the Taylor series of 1/(ζ − z) about z0 6= ζ is given by
∞
X (z − z0 )n
1
=
ζ −z
(ζ − z0 )n+1
n=0
for |z − z0 | < |ζ − z0 |.
6. Obtain the Taylor series about z = i for the following function and state where the
series converges.
1+z
f (z) =
.
1−z
7. Let
f (z) =
tan z
(z + 1)(z 2 + 4)
2
What is the radius of convergence of the following series.
(a) The Maclaurin series.
(b) The Taylor series about the point z0 = 1.
(c) The Taylor series about the point z0 = π(1 + i).
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
2
8. Obtain the first 3 non-zero terms in the Maclaurin expansion of the following stating
in each case where the series converges.
(a)
ez
.
z−1
(b)
tanh z =
sinh z
.
cosh z
9. By any means determine the radius of convergence of the following power series
1 + 2z + z 2 + (2z)3 + z 4 + (2z)5 + · · · + z 2n + (2z)2n+1 + · · ·
Give an expression for the limit.
10. Find the general term in the Maclaurin series of the following function.
∞
X
ez
=
cn z n .
1−z
n=0
11. Show that the following function is an entire function.
(
sin z , if z 6= 0,
z
f (z) =
1,
if z = 0.
12. Let
f (z) =
∞ 3
X
n
n=0
3n
zn.
(a) Determine where the series converges.
(b) Determine f (6) (0).
(c) Let C denote the unit circle traversed once in the anti-clockwise direction.
Determine
I
I
f (z) sin z
f (z)
and
dz.
4 dz
z2
C
C z
13. Find the Laurent series for the function
z
(z + 1)(z − 2)
in each of the following domains.
(a) |z| < 1.
(b) 1 < |z| < 2.
(c) 2 < |z|.
14. Find the Laurent series for
(z + 1)
z(z − 4)3
in 0 < |z − 4| < 4.
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
3
15. Find the first three non-zero terms of the Laurent series for each of the following
functions in the specified domains.
(a)
e1/z
,
z2 − 1
|z| > 1.
(b)
cosec(z) =
1
,
sin(z)
0 < |z| < π.
16. Determine the annulus of convergence of the Laurent series
∞
X
zj
.
2|j|
j=−∞
17. The following was question 3 of the May 2014 MA3614 exam paper.
(a) Determine if the following power series define entire functions and if this is not
the case then find the circle of convergence. In each case you must justify your
answer.
i.
∞
X
2n
n=0
n!
(z − 3)n .
[2 marks]
ii.
∞
X
(z + 1)n
n=2
n−1
.
[2 marks]
(b) Let
1 + sin z
.
1 + ez
Using knowledge of the functions sin z and ez , determine the circle of convergence of the Maclaurin series of g(z).
g(z) =
[2 marks]
Suppose that the Maclaurin series of g(z) is expressed in the form
g(z) = a0 + a1 z + a2 z 2 + a3 z 3 + · · · + an z n + · · · .
By writing the relation in the form
g(z)(1 + ez ) = 1 + sin z
determine a0 , a1 , a2 and a3 .
[6 marks]
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
4
(c) Let
1
.
(z + 1)(z + 3)
Determine each of the following three series:
f (z) =
(i) The power series of f (z) in |z| < 1.
(ii) The Laurent series of f (z) in 1 < |z| < 3.
(iii) The Laurent series of f (z) in |z| > 3.
[8 marks]
18. This was question 3a of the 2013 MA3914 paper.
Find the circle of convergence of the following power series justifying your answer
in each case.
(a)
∞
X
4n (z − 1)n .
n=0
[2 marks]
(b)
∞ 2
X
n
n=0
2n
zn.
[2 marks]
(c)
∞
X
n=1
an z n
(
n, if n is odd,
where an =
n2 , if n is even.
[3 marks]
19. This was question 3b of the 2013 MA3914 paper.
(a) Determine the Maclaurin series for z cos z and indicate where it converges.
[2 marks]
(b) Determine the Laurent series of
cos z
z2
about the point z = 0 and indicate where it converges.
[3 marks]
(c) Find the first 3 non-zero terms of the Laurent series of
z
sin z
in a region of the form 0 < |z| < R. State the largest value of R for which the
Laurent series converges and give a reason to justify your answer.
[4 marks]
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
5
(d) The function
f (z) =
3
(z + 1)(2z − 1)
is analytic in the annulus {z : 1/2 < |z| < 1}. Determine the Laurent series
for this function giving the general term involving z n for n ≥ 0 and for n < 0.
[4 marks]
20. This was question 4a of the 2011 MA3914 paper.
Determine the radius of convergence R of the following power series
f (z) =
∞
X
aj z j
with aj =
j=0
1
− 1.
(3i)j+1
Using the geometric series, show that the above power series can be re-summed
inside the radius of convergence |z| < R to give the function
f (z) =
1 − 3i
.
(z − 1)(z − 3i)
Defining the above f (z) as a function in the entire complex plane, determine in
which regions it can be expanded into a Laurent series and sketch these regions
including all the poles of f (z).
Perform a Laurent expansion
f (z) =
∞
X
cj z j
j=−∞
in the region containing the point z0 = 2, giving the coefficients cj explicitly.
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
6
Answers
1. Find the circle of convergence of the following power series justifying your answer
in each case.
(a)
∞
X
(z − i)n
n=0
2n
.
Answer
This is a geometric series with common ratio
z−i
.
2
The series converges for
z − i
2 <1
and diverges for
z − i
2 >1
and the circle of convergence is
|z − i| = 2.
(b)
∞
X
(z + 4i)2n (n − 1)2 .
n=0
Answer
Let
bn = (z + 4i)2n (n − 1)2 .
Use the ratio test.
bn+1 (z + 4i)2 n2 2
bn = (n − 1)2 → |z + 4i|
as
n2
=
(n − 1)2
1
1 − (1/n)
as n → ∞
2
→ 1 as n → ∞.
The circle of convergence is |z + 4i| = 1.
2. Suppose that a pseudo random number generator generates a sequence a0 , a1 , · · ·
with |an | ≤ 1. Find the circle of convergence of the power series
∞
X
n=0
(2 + an )z n .
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
7
Answer
Let
bn = (2 + an )z n .
The ratio test does not help here on its own as we can only deduce that |bn+1 /bn |
remains bounded. However with the root test we have
|bn |1/n = |2 + an |1/n |z|.
As
1 ≤ |2 + an | ≤ 3
and
1 ≤ |2 + an |1/n ≤ 31/n → 1.
Thus
|bn |1/n → |z| as n → ∞.
By the root test the circle of convergence is |z| = 1.
3. Obtain the Maclaurin series for z 3 sin(2z) and state where it converges.
Answer
The Maclaurin series for sin(z) is
sin(z) = z −
z 2n+1
z3 z5 z7
+
−
+ · · · + (−1)n
+ ···
3!
5!
7!
(2n + 1)!
The Maclaurin series for sin(2z) is
sin(2z) = (2z) −
(2z)3 (2z)5 (2z)7
(2z)2n+1
+
−
+ · · · + (−1)n
+ ···
3!
5!
7!
(2n + 1)!
The Maclaurin expansion for f (z) = z 3 sin(2z) is
8
32 8
22n+1 2n+4
z + · · · + +(−1)n
z
+ ···
f (z) = 2z 4 − z 6 +
6
120
(2n + 1)!
4. Let
∞
X
an z n
n=0
be the Maclaurin series of f (z) analytic at the origin. Show that
∞
X
0
is the Maclaurin series of g(z) = f (z 2 ).
an z 2n
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
8
Answer
Let
g(z) = f (z 2 ) = b0 + b1 z + b2 z 2 + · · · + bn z n + · · ·
As g(z) is an even function it follows by using the generalised Cauchy integral
formula, as is done in the lecture notes, that it only involves even powers, i.e.
0 = b1 = b3 = · · · = b2n+1 = · · · .
The generalised Cauchy integral formula gives an expression for the coefficients as
I
I
1
1
g (m) (0)
g(z)
f (z 2 )
=
dz
=
dz
bm =
m!
2πi C z m+1
2πi C z m+1
where C is a circle of some radius r with centre at 0. With z = r eit as a parametrization we have
Z 2π
1
f (r2 e2it )e−mit dt.
bm =
2πirm 0
If m is even, i.e. m = 2n, then
b2n
1
=
2πir2n
Z
2π
f (r2 e2it )e−2nit dt.
0
If we let s = 2t then as f (r2 eis )e−ins is 2π-periodic we have
Z 4π
Z 2π
1
1
1
2 is −nis
b2n =
f (r e )e
ds =
f (r2 eis )e−nis ds = an
2 2πir2n 0
2πir2n 0
as we have generalised Cauchy integral formula for an using a circle of radius r2 .
5. Prove that the Taylor series of 1/(ζ − z) about z0 6= ζ is given by
∞
X (z − z0 )n
1
=
ζ −z
(ζ − z0 )n+1
n=0
for |z − z0 | < |ζ − z0 |.
Answer
z − z0
ζ − z = (ζ − z0 ) − (z − z0 ) = (ζ − z0 ) 1 −
.
ζ − z0
When
z − z0 ζ − z0 < 1
we have the geometric series
(ζ − z)−1 = (ζ − z0 )−1
=
1+
∞
X
(z − z0 )n
n+1 .
(ζ
−
z
)
0
n=0
z − z0
ζ − z0
+
z − z0
ζ − z0
2
+
z − z0
ζ − z0
!
3
+ ···
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
9
6. Obtain the Taylor series about z = i for the following function and state where the
series converges.
1+z
.
f (z) =
1−z
Answer
√
The only point where f (z) is not analytic is z = 1 and as this is a distance
|1−i|
=
2
√
from z = i the Taylor series about z = i converges in |z − i| < 2 and diverges for
all other z.
To get the series it helps to first simplify f (z) by writing
1 + z = (−1)(1 − z) + 1,
f (z) = −1 +
1
.
1−z
Repeatedly differentiating f (z) is straightforward in this form as
1
,
(1 − z)2
···
···
n!
(n)
f (z) =
.
(1 − z)n+1
f 0 (z) =
Evaluating the derivatives at z = i leads to the Taylor series
f (n) (i)
f 00 (i)
2
(z − i) + · · · +
a(z − i)n + · · ·
f (z) = f (i) + f (i)(z − i) +
2
n!
z−i
1
(z − i)n
+
= −1 +
+ ··· +
+ ···
1−i
(1 − i)2
(1 − i)n+1
0
An alternative derivation is to write
z−i
1 − z = (1 − i) − (z − i) = (1 − i) 1 −
1−i
and hence by the geometric series
1+z
= −1 +
1−z
1
1−i
1+
z−i
1−i
7. Let
f (z) =
+
z−i
1−i
2
+ ··· +
tan z
(z + 1)(z 2 + 4)
2
What is the radius of convergence of the following series.
(a) The Maclaurin series.
(b) The Taylor series about the point z0 = 1.
(c) The Taylor series about the point z0 = π(1 + i).
z−i
1−i
!
n
+ ···
.
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
10
Answer
(a) The function has isolated singularities at ±i, ±2i and at the equally spaced
points π/2 + kπ, k ∈ Z.
The nearest singularities to the origin are at ±i and thus the radius of convergence of Maclaurin’s series is 1.
(b) The nearest singularity to z0 = 1 is at π/2 and thus the radius of convergence
is π/2 − 1.
(c) When z0 = π(1 + i),
|z0 − 2i|2 = |π + (π − 2)i|2 = π 2 + (π − 2)2
and
|z0 − π/2|2 = |π/2 + πi|2 = (π/2)2 + π 2 .
As π/2 >pπ − 2 the nearest singularity to z0 is at 2i The radius of convergence
is hence π 2 + (π − 2)2 .
8. Obtain the first 3 non-zero terms in the Maclaurin expansion of the following stating
in each case where the series converges.
(a)
ez
.
z−1
(b)
tanh z =
sinh z
.
cosh z
Answer
(a) Let
f (z) = a0 + a1 z + a2 z 2 + · · ·
(z − 1)f (z) = ez
gives
z2 z3
+
+ ···
2
6
We equate the coefficients of the powers of z on the left hand side with the
right hand side.
Constant term: −a0 = 1. Thus a0 = −1.
z term: −a1 + a0 = 1. This gives a1 = −2.
z 2 term: −a2 + a1 = 1/2. This gives a2 = a1 − 1/2 = −2 − 1/2 = −5/2.
Thus
5
f (z) = −1 − 2z − z 2 + · · ·
2
The only point where f (z) is not analytic is z = 1 and thus the Maclaurin
series converges for |z| < 1.
(z − 1)(a0 + a1 z + a2 z 2 + · · · ) = 1 + z +
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
11
(b) As sinh z is an odd function and cosh z is an even function the function tanh z
is an odd function and the Maclaurin series only involves odd powers and is of
the form
tanh z = a1 z + a3 z 3 + a5 z 5 + · · ·
As
(tanh z)(cosh z) = sinh z
we have
z3
z5
z2 z4
+
+ ··· = z +
+
+ ···
(a1 z + a3 z + a5 z + · · · ) 1 +
2
24
6
120
3
5
We equate the coefficients of the powers of z on the left hand side with the
right hand side.
z term: a1 = 1.
z 3 term: a3 + a1 /2 = 1/6. This gives a3 = 1/6 − 1/2 = −1/3.
z 5 term: a5 + a3 /2 + a1 /24 = 1/120. This gives
1
1
1 + 20 − 5
16
2
1
+ −
=
=
= .
a5 =
120 6 24
120
120
15
tanh z is not analytic when cosh z = 0 and
1
e−z 2z
cosh z = (ez + e−z ) =
(e + 1).
2
2
The points where tanh z is not analytic which are nearest to z = 0 are when
e2z = −1, i.e. z = ±iπ/2. The Maclaurin series converges for |z| < π/2.
9. By any means determine the radius of convergence of the following power series
1 + 2z + z 2 + (2z)3 + z 4 + (2z)5 + · · · + z 2n + (2z)2n+1 + · · ·
Give an expression for the limit.
Answer
The ratio test and the root test both do not work in their standard form to determine
the radius of convergence but we can consider the even powers and the odd powers
separately as these are just geometric series. The even powers gives
1
, |z| < 1.
1 + z2 + z4 + · · · + · · · =
1 − z2
The odd powers gives
2z
2z(1 + (2z)2 + (2z)4 + · · · ) =
, |2z| < 1.
1 − (2z)2
Both series are valid when |z| < 1/2 and thus the radius of convergence is 1/2.
The limit function is
1
2z
.
2 +
1−z
1 − (2z)2
10. Find the general term in the Maclaurin series of the following function.
∞
X
ez
=
cn z n .
1−z
n=0
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
12
Answer
The Maclaurin series for ez and 1/(1 − z) are standard and we have
∞
X
ez
z2
zn
2
n
= 1+z+
+ ··· +
+ ···
1 + z + z + ··· + z + ··· =
cn z n .
1−z
2!
n!
n=0
By the Cauchy product result we have
n
X
1
cn =
.
k!
k=0
The series is valid for |z| < 1.
11. Show that the following function is an entire function.
(
sin z , if z 6= 0,
z
f (z) =
1,
if z = 0.
Answer
We have the Laurent series
∞
X
z2 z4
sin z
z 2n
n
=1−
+
+ ··· =
(−1)
z
3!
5!
(2n + 1)!
n=0
valid for 0 < |z| < ∞. The right hand side is 1 when z = 0. Hence f (z) has the
Maclaurin series
∞
X
z 2n
z2 z4
(−1)n
+
+ ··· =
f (z) = 1 −
3!
5!
(2n + 1)!
n=0
which is valid for all z, i.e. the radius of convergence is ∞, and thus f (z) is an
entire function.
12. Let
f (z) =
∞ 3
X
n
n
n=0
3
zn.
(a) Determine where the series converges.
Let
3
n
.
an =
3n
We use the ratio test.
3 3 an+1
n+1
1
1
1
1
=
= 1+
→
an
n
3
n
3
3
as n → ∞. From this we have R = 3 as the radius of convergence. The series
converges for |z| < 3 and it converges uniformly in closed disks of the form
|z| ≤ R0 < 3.
Answer
(b) Determine f (6) (0).
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
Answer
The coefficient of z 6 is
13
f (6) (0)
63
= a6 = 6 .
6!
3
Thus
f
(6)
(0) = 6!a6 = 6!
63
36
.
(c) Let C denote the unit circle traversed once in the anti-clockwise direction.
Determine
I
I
f (z)
f (z) sin z
and
dz.
4 dz
z2
C z
C
Answer
The ratio has the Laurent series
a1 a2 a3
f (z)
+ a4 + a5 z + · · ·
4 = 3 + 2 +
z
z
z
z
in 0 < |z| < 3. The circle C is in this region and thus by term-by-term
integration we have
I
f (z)
4 dz = (2πi)a3 = 2πi.
C z
f (z) sin z =
z 8z 2
+
···
3
9
z3
z2
z−
+ ··· =
+ ···
6
3
and we have the Laurent series
1
f (z) sin z
= + ···
2
3
z
There is no term involving z −1 and thus
I
f (z) sin z
dz = 0.
z2
C
13. Find the Laurent series for the function
z
(z + 1)(z − 2)
in each of the following domains.
(a) |z| < 1.
(b) 1 < |z| < 2.
(c) 2 < |z|.
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
14
Answer
It helps here to first express the rational function in partial fraction form. We have
f (z) =
A
B
z
=
+
.
(z + 1)(z − 2)
z+1 z−2
and
z = A(z − 2) + B(z + 1).
By letting z = 2 we get 2 = 3B. By letting z = −1 we get −1 = −3A. Thus
2
1
1
1
f (z) =
+
.
3 z−2
3 z+1
Each of the terms can be represented by a power series inside a circle and by a
Laurent series only involving negative powers outside of a circle.
Now one of the easier formula to remember is the geometric series result
1
= 1 + α + α2 + · · · ,
1−α
|α| < 1
and in the following we put things in a form where this can be used.
For the power series
z
z − 2 = −2 1 −
2
1
1
z −1
= − 1−
z−2
2
2
z z 2 z 3
1
1+
+
+
+ ··· ,
= −
2
2
2
2
1
= 1 − z + z 2 − z 3 + · · · , |z| < 1.
z+1
|z| < 2,
For the series valid outside of a circle we make the substitution w = 1/z so that we
have
1
1 − 2w
−2=
,
w
w
w
1
=
,
z−2
1 − 2w
= w 1 + (2w) + (2w)2 + · · · , |2w| < 1,
!
2
1
2
2
1+
+
+ · · · , |z| > 2,
=
z
z
z
z−2 =
1
1+w
+1=
,
w
w
1
w
=
z+1
1+w
= w 1 − w + w2 − w3 + · · · , |w| < 1,
!
2 3
1
1
1
1
=
1− +
−
+ · · · , |z| > 1.
z
z
z
z
z+1 =
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
15
In the case of (a) we combine the two power series to give
X
n ∞ 1
1
n
f (z) =
(−1) −
zn.
3 n=0
2
In the case of (b) we need the power series for the B/(z − 2) term and the series for
A/(z + 1) valid for |z| > 1 which gives
X
X
∞
∞ n
(−1)n−1
1
1
1
−
zn.
f (z) =
n
3
z
3
2
1
0
In the case of (c) it is both series just involving negative powers.
X
∞
1 n
1
n
n−1
2 + (−1)
.
f (z) =
3
z
1
14. Find the Laurent series for
(z + 1)
z(z − 4)3
in 0 < |z − 4| < 4.
Answer
Let w = z − 4 so that z = w + 4.
w+5
A
B
C
D
(z + 1)
+ + 2 + 3.
3 =
3 =
w+4 w w
z(z − 4)
(w + 4)w
w
To get the constants A, B, C and D we have
f (z) =
w + 5 = Aw3 + (w + 4)(Bw2 + Cw + D).
Letting w = −4 gives A = −1/64.
Letting w = 0 gives 5 = 4D, D = 4/5.
w+5
1
Aw3
=1+
= D + Cw + Bw2 +
.
w+4
w+4
w+4
Differentiating gives
−
1
= C + 2Bw + O(w2 ),
(w + 4)2
C=−
1
.
16
Differentiating again gives
2
= 2B + O(w),
(w + 4)3
B=
1
.
64
The Laurent series is
∞
n
(5/4) (1/16) (1/64)
1 X
n w
f (z) =
−
+
−
(−1)
.
w
256 n=0
4
w3
w2
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
16
15. Find the first three non-zero terms of the Laurent series for each of the following
functions in the specified domains.
(a)
e1/z
,
z2 − 1
|z| > 1.
Answer
From the Maclaurin series
ew = 1 + w +
w
wn
+ ··· +
+ ···
2!
n!
we get the Laurent series
e1/z = 1 +
1 1
1 1
1
+
+ ···
2 + ··· +
z 2! z
n! z n
which is valid for |z| > 0.
Let now w = 1/z with |z| > 1.
2
1 − w2
1
−1=
,
z −1=
w
w2
2
giving
1
w2
=
= w2 (1 + w2 + w4 + · · · ) =
z2 − 1
1 − w2
2 4 6
1
1
1
+
+
+ ···
z
z
z
Hence
!
2 4 6
1
1 1
1
1
1
1+ +
+ ···
+
+
+ ···
z 2! z 2
z
z
z
2 3 4
1
1
1
1
+
+ 1+
+ ···
=
z
z
2
z
2 3
4
1
1
3 1
=
+
+
+ ···
z
z
2 z
e1/z
=
z2 − 1
(b)
cosec(z) =
1
,
sin(z)
0 < |z| < π.
Answer
sin z has a simple zero at z = 0 and it is an odd function. Thus the series is
of the form
1
a−1
f (z) =
=
+ a1 z + a3 z 3 + · · · .
sin z
z
1 = f (z) sin z
a
z3
z5
−1
3
=
+ a1 z + a3 z + · · ·
z−
+
+ ··· .
z
6
120
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
17
Equating the constant term gives a−1 = 1.
Equating the coefficient of z 2 gives
a1 −
a−1
= 0,
6
1
a1 = .
6
Equating the coefficient of z 4 gives
a3 −
a1 a−1
+
= 0,
6
120
a3 =
1
1
7
−
=
.
36 120
360
16. Determine the annulus of convergence of the Laurent series
∞
X
zj
.
|j|
2
j=−∞
Answer
−1
∞
X
X
zj
1
=
j.
|j|
(2z)
2
j=−∞
j=1
This converges when |2z| > 1.
∞
∞
X
X
zj
zj
=
j.
|j|
2
2
j=0
j=0
This converges when |z/2| < 1.
The annulus of convergence is
1
< |z| < 2.
2
17. The following was question 3 of the May 2014 MA3614 exam paper.
(a) Determine if the following power series define entire functions and if this is not
the case then find the circle of convergence. In each case you must justify your
answer.
i.
∞
X
2n
n=0
n!
(z − 3)n .
[2 marks]
ii.
∞
X
(z + 1)n
n=2
n−1
.
[2 marks]
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
18
(b) Let
1 + sin z
.
1 + ez
Using knowledge of the functions sin z and ez , determine the circle of convergence of the Maclaurin series of g(z).
g(z) =
[2 marks]
Suppose that the Maclaurin series of g(z) is expressed in the form
g(z) = a0 + a1 z + a2 z 2 + a3 z 3 + · · · + an z n + · · · .
By writing the relation in the form
g(z)(1 + ez ) = 1 + sin z
determine a0 , a1 , a2 and a3 .
[6 marks]
(c) Let
1
.
(z + 1)(z + 3)
Determine each of the following three series:
f (z) =
(i) The power series of f (z) in |z| < 1.
(ii) The Laurent series of f (z) in 1 < |z| < 3.
(iii) The Laurent series of f (z) in |z| > 3.
[8 marks]
Answer
(a)
i. Let
an =
2n
(z − 3)n .
n!
Using the ratio test
an+1
2
(z − 3) → 0 as n → ∞
=
an
n+1
for all z. The function is an entire function.
2 marks
ii. Let
an =
(z + 1)n
.
n−1
Using the ratio test
an+1 (z + 1)(n − 1) =
→ |z + 1| as n → ∞.
an n
The series converges for |z + 1| < 1 and diverges for |z + 1| > 1. The circle
of convergence is {z : |z + 1| = 1}.
2 marks
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
19
(b) sin z and ez are entire functions and thus the only points where f (z) is not
analytic are points for which 1 + ez = 0. The nearest such points to z = 0 are
±iπ and thus the circle of convergence is {z : |z| = π}.
2 marks
Using the series for ez and sin z we have
z2 z3
z3
z5
2
3
(a0 +a1 z+a2 z +a3 z +· · · ) 2 + z +
+
+ ··· = 1 + z −
+
+ ··· .
2
6
6
120
We can equate the coefficients of the powers.
z 0 terms: 2a0 = 1, a0 = 1/2.
z 1 terms: a0 + 2a1 = 1/2 + 2a1 = 1, a1 = 1/4.
z 2 terms: a0 /2 + a1 + 2a2 = 1/4 + 1/4 + 2a2 = 1/2 + 2a2 = 0, a2 = −1/4.
z 3 terms: a0 /6+a1 /2+a2 +2a3 = 1/12+1/8−1/4+2a3 = −1/12+2a3 = −1/6,
a3 = −1/24.
6 marks
(c) By partial fractions we have
f (z) =
A
B
1
=
+
.
(z + 1)(z + 3)
z+1 z+3
1 = A(z + 3) + B(z + 1),
gives A = 1/2,
B = −1/2.
By the geometric series
1
= 1 − z + z 2 + · · · (−1)n z n + · · · ,
1+z
1
1/3
=
3+z
1 + (z/3)
1
=
1 − (z/3) + (z/3)2 + · · · (−1)n (z/3)n + · · · .
3
The general term of the power series of f (z) valid in |z| < 1 is
1
n1
(−z)
1 − n+1 .
2
3
Let w = 1/z and write
1
1
w
=
=
.
1+z
1 + 1/w
1+w
When |z| > 1, |w| < 1 and we have
1
w
=
= w − w2 + w3 + · · · (−1)n−1 wn + · · ·
1+z
1+w
= z −1 − z −2 + z −3 + · · · + (−1)n−1 z −n + · · · .
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
20
Similarly by writing w = 3/z this satisfies |w| < 1 when |z| > 3.
1
1/z
1
w
=
=
3+z
1 + 3/z
3 1+w
1
=
w − w2 + w3 + · · · + (−1)n−1 wn + · · ·
3
1
=
3z −1 − 32 z −2 + · · · + (−1)n−1 3n z −n + · · ·
3
= z −1 − 3z −2 + · · · + (−1)n−1 3n−1 z −n + · · ·
The Laurent series valid in |z| > 3 is
1
2
z −1 − z −2 + · · · (−1)n−1 z −n + · · ·
− z −1 − 3z −2 + · · · + (−1)n−1 3n−1 z −n + · · ·
.
Only negative powers are involved, the first term is 1/z 2 and the coefficient of
z −n is
1
(−1)n−1 1 − 3n−1 .
2
The Laurent series valid in 1 < |z| < 3 involves combining the power series for
the B/(3 + z) term with the Laurent series for the A/(1 + z) term. Hence we
have
1
− 1 − (z/3) + (z/3)2 + · · · (−1)n (z/3)n + · · ·
6
1 −1
z − z −2 + z −3 + · · · (−1)n−1 z −n + · · · .
+
2
8 marks
18. This was question 3a of the 2013 MA3914 paper.
Find the circle of convergence of the following power series justifying your answer
in each case.
(a)
∞
X
4n (z − 1)n .
n=0
[2 marks]
(b)
∞ 2
X
n
n
n=0
2
zn.
[2 marks]
(c)
∞
X
n=1
an z n
(
n, if n is odd,
where an =
n2 , if n is even.
[3 marks]
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
21
Answer
(a) This is a geometric series which converges if and only if
|4(z − 1)| < 1.
The circle of convergence is {z : |z − 1| = 1/4}.
2 marks
(b) Use the ratio test. Let
bn =
n2
2n
n
z ,
|bn+1 |
=
giving
|bn |
n+1
n
2
|z|
|z|
→
2
2
as n → ∞.
The series converges if |z| < 2 and diverges if |z| > 2 and thus the circle of
convergence is {z : |z| = 2}.
2 marks
(c) Use the root test. Let
b n = an z n ,
giving |bn |1/n = |an |1/n |z|.
As
n1/n ≤ |an |1/n ≤ n2/n
and the left hand side and right hand side both converge to 1 as n → ∞ it
follows that
|bn |1/n → |z| as n → ∞
and the series converges if |z| < 1 and diverges for |z| > 1 and thus the circle
of convergence is {z : |z| = 1}.
3 marks
19. This was question 3b of the 2013 MA3914 paper.
(a) Determine the Maclaurin series for z cos z and indicate where it converges.
[2 marks]
(b) Determine the Laurent series of
cos z
z2
about the point z = 0 and indicate where it converges.
[3 marks]
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
22
(c) Find the first 3 non-zero terms of the Laurent series of
z
sin z
in a region of the form 0 < |z| < R. State the largest value of R for which the
Laurent series converges and give a reason to justify your answer.
[4 marks]
(d) The function
f (z) =
3
(z + 1)(2z − 1)
is analytic in the annulus {z : 1/2 < |z| < 1}. Determine the Laurent series
for this function giving the general term involving z n for n ≥ 0 and for n < 0.
[4 marks]
Answer
(a) The Maclaurin series for cos z is a standard series and is given by
cos z = 1 −
Thus
z2 z4
z 2n
+
+ · · · + (−1)n
+ ···
2
24
(2n)!
2n+1
z3 z5
nz
z cos z = z −
+
+ · · · + (−1)
+ ···
2
24
(2n)!
and this converges for all z.
2 marks
(b) Using the series for cos z the Laurent series is
2n−2
cos z
1
1 z2
nz
=
−
+
+
·
·
·
+
(−1)
+ ···
(2n)!
z2
z 2 2 24
and this converges in {z : 0 < |z| < ∞}.
3 marks
(c) As the zeros of sin z are on the real axis and these are at kπ where k is an
integer it follows that the given function is analytic in 0 < |z| < R with R = π.
The function is bounded as z → 0 and thus it has a removable singularity at
z = 0. It is an even function and hence only even powers are involved and
hence for 0 < |z| < π
z
= a0 + a2 z 2 + a4 z 4 + · · · .
sin z
Rearranging and using the known series for sin z gives
z3
z5
2
4
z = (a0 + a2 z + a4 z + · · · ) z −
+
+ ··· .
6
120
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
23
Equating the coefficient of z gives a0 = 1.
Equating the coefficient of z 3 gives
a0
1
0 = a2 − ,
a2 = .
6
6
5
Equating the coefficient of z gives
a2
a0
a2
a0
1
1
7
0 = a4 −
+
,
a4 =
−
=
−
=
.
6
120
6
120
36 120
360
4 marks
(d)
3
A
B
=
+
.
(z + 1)(2z − 1)
z + 1 2z − 1
1 = A(2z − 1) + B(z + 1), 3 = −3A, 3 = 3B/2
f (z) =
and thus A = −1 and B = 2.
For the non-negative powers we have the geometric series
−1
= −1 + z − z 2 + z 3 + · · · + (−1)n+1 z n + · · ·
z+1
which is valid for |z| < 1.
Now to get the negative powers we write
1
2z − 1 = 2z 1 −
2z
giving
−1
2
1
2
1
1
1
=
1−
=2
+
+ ··· + n n + ···
2z − 1
2z
2z
2z 4z 2
2 z
which is valid for |2z| > 1. The Laurent series valid in the annulus is the sum
of the two series.
4 marks
20. This was question 4a of the 2011 MA3914 paper.
Determine the radius of convergence R of the following power series
∞
X
1
f (z) =
aj z j with aj =
j+1 − 1.
(3i)
j=0
Using the geometric series, show that the above power series can be re-summed
inside the radius of convergence |z| < R to give the function
1 − 3i
f (z) =
.
(z − 1)(z − 3i)
Defining the above f (z) as a function in the entire complex plane, determine in
which regions it can be expanded into a Laurent series and sketch these regions
including all the poles of f (z).
Perform a Laurent expansion
f (z) =
∞
X
cj z j
j=−∞
in the region containing the point z0 = 2, giving the coefficients cj explicitly.
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
24
Answer
∞ X
z n 1
n=0
3i
3i
Also
=
∞
X
1
3i
zn =
n=0
1−
1
z =
1
.
3i − z
3i
1
.
1−z
The function f (z) is
f (z) =
1
1
1
(z − 3i) − (z − 1)
1 − 3i
−1
+
=
−
=
=
.
1 − z 3i − z
z − 1 z − 3i
(z − 1)(z − 3i)
(z − 1)(z − 3i)
The function has simple poles at z = 1 and at z = 3i and thus it has a power series
in |z| < 1, a Laurent series involving positive and negative powers in 1 < |z| < 3
and a Laurent series involving only negative powers in |z| > 3.
The region 1 < |z| < 3 is shown next.
3i
1
The point z = 2 is in the annulus. Now
z
3i − z = (3i) 1 −
3i
which gives
1
=
3i − z
which is valid for |z| < 3.
1
3i
z z 2
1+
+
+ ···
3i
3i
c M. K. Warby
2-3-2015 14:26 MA3614 Complex variable methods and applications
If we let w = 1/z then
1−z =1−
Thus
1
w−1
=
,
w
w
−1
w
=
= w + w2 + w3 + · · ·
1−z
1−w
1
−1
1
1
= + 2 + 3 + ···
1−z
z z
z
valid for |z| > 1.
Thus
f (z) =
∞
X
cn z n
n=−∞
with cn = 1 for n < 0 and for n ≥ 0 we have
n+1
1
cn =
.
3i
25