Section 6: Power Series
Examples and Problems
Example 1
[convergence of power series]
Use the Cauchy-Hadamard formula to determine the radius of convergence of the following
power series (draw the centre and radius of convergence in the complex plane):

(a)
 (3z  2 j )
n
n 0
z n2

n 0 (n  1)( n  2)

(b)
Solution
(a)
This series is not written in the usual form of a power series. We rewrite it as

3
n
( z  2 j / 3) n
n 0
We can now apply the Cauchy-Hadamard formula:
a
3n1
lim n1  lim n  3
n a
n 3
n
Thus the radius of convergence is R  1 / 3 . The series converges
for any z inside the circle of radius 1 / 3 and centre 2 j / 3
z  2 j /3
( z  2 j / 3  1 / 3 ), and diverges for any z outside this circle. The
Cauchy-Hadamard formula does not tell us about the convergence
of the series on the circle of convergence (we would have to use
some other test to find out the behaviour there).
(b)
Applying the formula to the series gives
a
(n  1)( n  2)
lim n1  lim
1
n a
n  (n  2)( n  3)
n
Thus the radius of convergence is R  1 . The series converges for
any z inside the circle of convergence z  1 (the unit circle).
z0
[note: it doesn’t matter that the exponent is n  2 and not n . We
could rewrite the series as

zm
and we get the same answer.]

m  2 ( m  1) m
Complex Variables Section 6 Examples & Problems - 1
Example 2
[Taylor Series]
Find the Taylor series of the function f ( z )  sin z about the point z o   / 2 and determine the
radius of convergence

1
f
(
z
)

an ( z  zo ) n , an  f (n) ( zo ) ]
[use the formula

n!
n 0
Solution
The derivatives are
f ( z )  sin z, f ( z )  cos z, f ( z )   sin z, f ( z )   cos z, f ( IV ) ( z )  sin z, 
so
f ( z o )  1, f ( z o )  0, f ( z o )  1, f ( z o )  0, f ( IV ) ( z o )  1, 
Thus
1
1
1
f ( z )  1  ( z   / 2) 2  ( z   / 2) 4  ( z   / 2) 6  
2!
4!
6!

1
  (1) n
( z   / 2) 2 n
(2n)!
n 0
The function f ( z )  sin z is analytic everywhere in the complex plane (it is “entire”), so we can
immediately say that the radius of convergence is R   (remember, there will be a singular
point on the radius of convergence, but this function doesn’t have any singular points).
Example 3
[Derivation of Power/Taylor Series using other methods]
Use the geometric series

g ( z)  1  z  z 2  z 3     z n 
n 0
1
1 z
to expand the function 1 / z about the point 1  j .
Solution
First of all, from the sketch we see that
the radius of convergence should be 2 ,
the distance between the centre and the
nearest (only) singular point.
The centre of the series is z o  1  j so
we’ll need to expand in powers of
( z  1  j) .
z  1 j
singular
point
centre
z0
To use the geometric series, we’ll have to
rewrite our function in the form
Complex Variables Section 6 Examples & Problems - 2
f ( z) 
1
?
, and we have to find the missing constants.

z 1  ?( z  1  j )
We have
1
1

z (1  j )  ( z  1  j )
1

(1  j )  ( z  1  j )
1

 z 1 j 
(1  j ) 1 
 1  j 

This is in the appropriate form so we can write
2
3

 z 1 j 
1
1  z  1  j  z  1  j 
f ( z)  

1







 1 j 
 1  j   1  j 
 z  1  j  1  j 



(1  j ) 1 

1 j 

1
z  1  j ( z  1  j) 2 ( z  1  j) 3





1  j (1  j ) 2
(1  j ) 3
(1  j ) 4
The geometric series converges for z  1 so our series converges for
f ( z) 
z 1 j
z 1 j
z 1 j
1
1
 1  z 1 j  2
1 j
1 j
2
as expected (the radius of convergence is the circle of centre 1  j and radius 2 .
[note: multiplying a series by a constant (other than 0) as we have here (see the term 1 /(1  j ) )
makes no difference to the convergence or divergence of the series]
Example 4
[Derivation of Power/Taylor Series using other methods]
Use the exponential series,

zn
z2 z3
z
e 
 1 z 
 
2! 3!
n 0 n!
e z 1
to expand the function 3 about the point z o  0 .
z
4
Solution
It looks like this function has a singularity at z  0 , but let’s see what happens. From the
exponential series we can write

4
z 4n
z 8 z 12
ez  
 1 z4  

2! 3!
n 0 n!
Complex Variables Section 6 Examples & Problems - 3
Thus

ez 1
z5 z9
z 4 n 1

z






2! 3!
z3
n  0 ( n  1)!
This is a valid Taylor series with centre z o  0 and radius of convergence R   .
4
[note (don’t worry too much about these notes):
(1) a point that looks like a singularity, like z o  0 here, but in fact is not is called a removable
singularity.
(2) How can one know for sure that the above series has infinite radius of convergence? Well,
rewrite it as
4

 z 4 z8 
ez 1
z 4n

z
1



z



2! 3! 
z3
n 0 ( n  1)!

The z term converges everywhere, so the convergence will depend on the bit inside the

brackets. The series n0 z n /( n  1)! converges for z   (by the Cauchy-Hadamard

a n z n has radius of convergence R, then
an z 2 n has a radius of convergence
R , from which it follows that our series
formula). Now it can be shown that if a series
the series


n 0

n 0
also has radius of convergence R   ]
Complex Variables Section 6 Examples & Problems - 4
Problem 1
Use the Cauchy-Hadamard formula to determine the radius of convergence of the following
power series (draw the centre and radius of convergence in the complex plane):

(a)
 n! z
n
[ANSWER: R  0 ]
n 0

(b)
( z  2) n 1
3 n
4
 (n  1)
n 0
[ANSWER: R  4 ]
Problem 2
[Taylor Series]

1 (n)
f ( z o ) , to obtain the Taylor series of the
n!
n 0
following functions (and determine their radius of convergence)
(a) cos z about z o   / 2
[ANSWER: z   ]
Use the formula f ( z )   a n ( z  z o ) n , a n 
(b) Ln(1  z ) about z o  0
[ANSWER: z  1]
Problem 3
[Derivation of Power/Taylor Series using other methods]
Expand the following functions in Taylor series (determine the circles of convergence and plot,
together with any singular points, in the complex plane)
(a) 1 /( z  j ) 2 about  2 j (use the binomial series)
[ANSWER:  1  2 j ( z  2 j )  3( z  2 j ) 2   , z  2 j  1 ]
(b)
z 2  2 z  11
about z  1 (use partial fractions, and the binomial and geometric series)
( z  3) 2 (2 z  1)
[ANSWER:  7 / 8  31( z  1) / 16  509( z  1) 2 / 128   , z  1  1 / 2 ]
(c)
( z  j) 2
about z   j (use the geometric series)
z j
[ANSWER:  12 j ( z  j ) 2  14 ( z  j ) 3  18 j ( z  j ) 4  , z  j  2
Complex Variables Section 6 Examples & Problems - 5