Taylor Series, Laurent Series - Complex Analysis, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


Taylor and Laurent Series
4.1. Taylor Series
4.1.1. Taylor Series for Holomorphic Functions. In Real Analysis, the Taylor
series of a given function f :R!R is given by:
f(x
0
)+ f
0
(x
0
)(x x
0
)+
f
00
(x
0
)
2!
(x x
0
)
2
+
f
000
(x
0
)
3!
(x x
0
)
3
+ . . .
We have examined some convergence issues and applications of Taylor series in MATH
2033/2043. We also learned that even if the function f is in?nitely differentiable
everywhere onR, its Taylor series may not converge to that function. In contrast,
there is no such an issue in Complex Analysis: as long as the function f :C!C is
holomorphic on an open ball B
d
(z
0
), we can show the Taylor series of f :
f(z
0
)+ f
0
(z
0
)(z z
0
)+
f
00
(z
0
)
2!
(z z
0
)
2
+
f
000
(z
0
)
3!
(z z
0
)
3
+ . . .
converges pointwise to f(z) on B
d
(z
0
), and uniformly on any smaller ball. As we shall
see, it thanks to Cauchy’s integral formula. Moreover, the proof of Taylor Theorem
in Complex Analysis is also much easier than that in Real Analysis, again thanks to
Cauchy’s integral formula.
In this chapter, it is more convenient to re-label the variables in the Cauchy’s
integral formula:
f
(n)
(a)=
n!
2pi
I
g
f(z)
(za)
n+1
dz ! f
(n)
(z)=
n!
2pi
I
g
f(x)
(x z)
n+1
dx.
For the re-labelled Cauchy’s integral formula, we require the point z to be enclosed by
the simple closed curve g.
Theorem 4.1 (Taylor Theorem for Holomorphic Functions). Given a complex-valued
function f which is holomorphic on an open ball B
R
(z
0
), the series:
¥
å
n=0
f
(n)
(z
0
)
n!
(z z
0
)
n
converges (pointwise) to f(z) for any z2 B
R
(z
0
).
Page 2


Taylor and Laurent Series
4.1. Taylor Series
4.1.1. Taylor Series for Holomorphic Functions. In Real Analysis, the Taylor
series of a given function f :R!R is given by:
f(x
0
)+ f
0
(x
0
)(x x
0
)+
f
00
(x
0
)
2!
(x x
0
)
2
+
f
000
(x
0
)
3!
(x x
0
)
3
+ . . .
We have examined some convergence issues and applications of Taylor series in MATH
2033/2043. We also learned that even if the function f is in?nitely differentiable
everywhere onR, its Taylor series may not converge to that function. In contrast,
there is no such an issue in Complex Analysis: as long as the function f :C!C is
holomorphic on an open ball B
d
(z
0
), we can show the Taylor series of f :
f(z
0
)+ f
0
(z
0
)(z z
0
)+
f
00
(z
0
)
2!
(z z
0
)
2
+
f
000
(z
0
)
3!
(z z
0
)
3
+ . . .
converges pointwise to f(z) on B
d
(z
0
), and uniformly on any smaller ball. As we shall
see, it thanks to Cauchy’s integral formula. Moreover, the proof of Taylor Theorem
in Complex Analysis is also much easier than that in Real Analysis, again thanks to
Cauchy’s integral formula.
In this chapter, it is more convenient to re-label the variables in the Cauchy’s
integral formula:
f
(n)
(a)=
n!
2pi
I
g
f(z)
(za)
n+1
dz ! f
(n)
(z)=
n!
2pi
I
g
f(x)
(x z)
n+1
dx.
For the re-labelled Cauchy’s integral formula, we require the point z to be enclosed by
the simple closed curve g.
Theorem 4.1 (Taylor Theorem for Holomorphic Functions). Given a complex-valued
function f which is holomorphic on an open ball B
R
(z
0
), the series:
¥
å
n=0
f
(n)
(z
0
)
n!
(z z
0
)
n
converges (pointwise) to f(z) for any z2 B
R
(z
0
).
Proof. Given any z2 B
R
(z
0
), we let # > 0 be small enough so thatjz z
0
j < R#.
For simplicity, denote R
0
= R#.
By Cauchy’s integral formula, for any z2 B
R
0(z
0
), we have:
f(z)=
1
2pi
I
jxz
0
j=R
0
f(x)
x z
dx.
Then, the contourjz z
0
j = R
0
lies inside the open ball B
R
(z
0
). The key trick to prove
the Taylor Theorem is rewriting
1
x z
as a geometric series. Recall that:
1
1 w
= 1+ w+ w
2
+ . . . wheneverjwj< 1.
We ?rst rewrite
1
x z
into this form:
1
x z
=
1
(x z
0
)(z z
0
)
=
1
x z
0


1
1
zz
0
xz
0
=
1
x z
0
¥
å
n=0

z z
0
x z
0

n
Here we have used the fact that



zz
0
xz
0


< 1. See the diagram below. The yellow ball is
B
R
(z
0
), and the red circle isjx z
0
j = R
0
.
z
0
x
z
jx z
0
j = R
0
jz z
0
j< R
0
Then, whenever z2 B
R
0(z
0
), the function f(z) can be expressed as:
f(z)=
1
2pi
I
jxz
0
j=R
0
f(x)

1
x z
dx (4.1)
=
1
2pi
I
jxz
0
j=R
0
f(x)
x z
0
¥
å
n=0

z z
0
x z
0

n
dx
=
1
2pi
I
jxz
0
j=R
0
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
dx
Next we want to see whether we can switch the integral sign
I
jxz
0
j=R
0
and the in?nite
summation
¥
å
n=0
. For this we need to show uniform convergence of the series below.
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
.
Page 3


Taylor and Laurent Series
4.1. Taylor Series
4.1.1. Taylor Series for Holomorphic Functions. In Real Analysis, the Taylor
series of a given function f :R!R is given by:
f(x
0
)+ f
0
(x
0
)(x x
0
)+
f
00
(x
0
)
2!
(x x
0
)
2
+
f
000
(x
0
)
3!
(x x
0
)
3
+ . . .
We have examined some convergence issues and applications of Taylor series in MATH
2033/2043. We also learned that even if the function f is in?nitely differentiable
everywhere onR, its Taylor series may not converge to that function. In contrast,
there is no such an issue in Complex Analysis: as long as the function f :C!C is
holomorphic on an open ball B
d
(z
0
), we can show the Taylor series of f :
f(z
0
)+ f
0
(z
0
)(z z
0
)+
f
00
(z
0
)
2!
(z z
0
)
2
+
f
000
(z
0
)
3!
(z z
0
)
3
+ . . .
converges pointwise to f(z) on B
d
(z
0
), and uniformly on any smaller ball. As we shall
see, it thanks to Cauchy’s integral formula. Moreover, the proof of Taylor Theorem
in Complex Analysis is also much easier than that in Real Analysis, again thanks to
Cauchy’s integral formula.
In this chapter, it is more convenient to re-label the variables in the Cauchy’s
integral formula:
f
(n)
(a)=
n!
2pi
I
g
f(z)
(za)
n+1
dz ! f
(n)
(z)=
n!
2pi
I
g
f(x)
(x z)
n+1
dx.
For the re-labelled Cauchy’s integral formula, we require the point z to be enclosed by
the simple closed curve g.
Theorem 4.1 (Taylor Theorem for Holomorphic Functions). Given a complex-valued
function f which is holomorphic on an open ball B
R
(z
0
), the series:
¥
å
n=0
f
(n)
(z
0
)
n!
(z z
0
)
n
converges (pointwise) to f(z) for any z2 B
R
(z
0
).
Proof. Given any z2 B
R
(z
0
), we let # > 0 be small enough so thatjz z
0
j < R#.
For simplicity, denote R
0
= R#.
By Cauchy’s integral formula, for any z2 B
R
0(z
0
), we have:
f(z)=
1
2pi
I
jxz
0
j=R
0
f(x)
x z
dx.
Then, the contourjz z
0
j = R
0
lies inside the open ball B
R
(z
0
). The key trick to prove
the Taylor Theorem is rewriting
1
x z
as a geometric series. Recall that:
1
1 w
= 1+ w+ w
2
+ . . . wheneverjwj< 1.
We ?rst rewrite
1
x z
into this form:
1
x z
=
1
(x z
0
)(z z
0
)
=
1
x z
0


1
1
zz
0
xz
0
=
1
x z
0
¥
å
n=0

z z
0
x z
0

n
Here we have used the fact that



zz
0
xz
0


< 1. See the diagram below. The yellow ball is
B
R
(z
0
), and the red circle isjx z
0
j = R
0
.
z
0
x
z
jx z
0
j = R
0
jz z
0
j< R
0
Then, whenever z2 B
R
0(z
0
), the function f(z) can be expressed as:
f(z)=
1
2pi
I
jxz
0
j=R
0
f(x)

1
x z
dx (4.1)
=
1
2pi
I
jxz
0
j=R
0
f(x)
x z
0
¥
å
n=0

z z
0
x z
0

n
dx
=
1
2pi
I
jxz
0
j=R
0
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
dx
Next we want to see whether we can switch the integral sign
I
jxz
0
j=R
0
and the in?nite
summation
¥
å
n=0
. For this we need to show uniform convergence of the series below.
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
.
We use Weiestrass’s M-test: for any x on the circlefjx z
0
j = R
0
g, we have:




f(x)(z z
0
)
n
(x z
0
)
n+1









(z z
0
)
n
(x z
0
)
n+1




sup
jxz
0
j=R
0
j f(x)j
| {z }
=:C
R
0
=
C
R
0
R
0

jz z
0
j
R
0

n
Sincejz z
0
j< R
0
, the series
¥
å
n=0
C
R
0
R
0

jz z
0
j
R
0

n
converges. Note that the above series does not depend on x (the integration variable).
Hence by Weiestrass’s M-test, the series
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
converges uniformly on
the circlefjx z
0
j = R
0
g, thus allowing the switch between the integral sign and the
summation sign in (4.1):
f(z)=
1
2pi
¥
å
n=0
I
jxz
0
j=R
0
f(x)(z z
0
)
n
(x z
0
)
n+1
dx
=
¥
å
n=0
1
2pi

I
jxz
0
j=R
0
f(x)
(x z
0
)
n+1
dx

(z z
0
)
n
=
¥
å
n=0
f
(n)
(z
0
)
n!
(z z
0
)
n
.
In the last step we have used the higher order Cauchy’s integral formula. 
Example 4.1. The function f(z)= sin z is an entire function. By straight-forward
computations, its derivatives are given by:
f
0
(z)= cos z f
00
(z)= sin z
f
(3)
(z)= cos z f
(4)
(z)= sin z
.
.
.
.
.
.
Inductively, it is easy to deduce that f
(2k+1)
(0)=(1)
k
, and f
(2k)
(0)= 0 for any
integer k 0. Hence, the Taylor series of f about 0 is given by:
f(z)=
¥
å
k=0
f
(2k+1)
(0)
(2k+ 1)!
z
2k+1
=
¥
å
k=0
(1)
k
(2k+ 1)!
z
2k+1
= z
z
3
3!
+
z
5
5!

z
7
7!
+ . . .
This series converges to sin z for any z2C, because sin z is entire (i.e. holomorphic
on every ball B
R
(0)).
Example 4.2. Consider the function f(z) = Log(z) which is holomorphic on
W :=Cnfx+ 0i : x 0g. Note that we can only apply Theorem 4.1 if the ball
B
R
(z
0
) is contained insideW.
Page 4


Taylor and Laurent Series
4.1. Taylor Series
4.1.1. Taylor Series for Holomorphic Functions. In Real Analysis, the Taylor
series of a given function f :R!R is given by:
f(x
0
)+ f
0
(x
0
)(x x
0
)+
f
00
(x
0
)
2!
(x x
0
)
2
+
f
000
(x
0
)
3!
(x x
0
)
3
+ . . .
We have examined some convergence issues and applications of Taylor series in MATH
2033/2043. We also learned that even if the function f is in?nitely differentiable
everywhere onR, its Taylor series may not converge to that function. In contrast,
there is no such an issue in Complex Analysis: as long as the function f :C!C is
holomorphic on an open ball B
d
(z
0
), we can show the Taylor series of f :
f(z
0
)+ f
0
(z
0
)(z z
0
)+
f
00
(z
0
)
2!
(z z
0
)
2
+
f
000
(z
0
)
3!
(z z
0
)
3
+ . . .
converges pointwise to f(z) on B
d
(z
0
), and uniformly on any smaller ball. As we shall
see, it thanks to Cauchy’s integral formula. Moreover, the proof of Taylor Theorem
in Complex Analysis is also much easier than that in Real Analysis, again thanks to
Cauchy’s integral formula.
In this chapter, it is more convenient to re-label the variables in the Cauchy’s
integral formula:
f
(n)
(a)=
n!
2pi
I
g
f(z)
(za)
n+1
dz ! f
(n)
(z)=
n!
2pi
I
g
f(x)
(x z)
n+1
dx.
For the re-labelled Cauchy’s integral formula, we require the point z to be enclosed by
the simple closed curve g.
Theorem 4.1 (Taylor Theorem for Holomorphic Functions). Given a complex-valued
function f which is holomorphic on an open ball B
R
(z
0
), the series:
¥
å
n=0
f
(n)
(z
0
)
n!
(z z
0
)
n
converges (pointwise) to f(z) for any z2 B
R
(z
0
).
Proof. Given any z2 B
R
(z
0
), we let # > 0 be small enough so thatjz z
0
j < R#.
For simplicity, denote R
0
= R#.
By Cauchy’s integral formula, for any z2 B
R
0(z
0
), we have:
f(z)=
1
2pi
I
jxz
0
j=R
0
f(x)
x z
dx.
Then, the contourjz z
0
j = R
0
lies inside the open ball B
R
(z
0
). The key trick to prove
the Taylor Theorem is rewriting
1
x z
as a geometric series. Recall that:
1
1 w
= 1+ w+ w
2
+ . . . wheneverjwj< 1.
We ?rst rewrite
1
x z
into this form:
1
x z
=
1
(x z
0
)(z z
0
)
=
1
x z
0


1
1
zz
0
xz
0
=
1
x z
0
¥
å
n=0

z z
0
x z
0

n
Here we have used the fact that



zz
0
xz
0


< 1. See the diagram below. The yellow ball is
B
R
(z
0
), and the red circle isjx z
0
j = R
0
.
z
0
x
z
jx z
0
j = R
0
jz z
0
j< R
0
Then, whenever z2 B
R
0(z
0
), the function f(z) can be expressed as:
f(z)=
1
2pi
I
jxz
0
j=R
0
f(x)

1
x z
dx (4.1)
=
1
2pi
I
jxz
0
j=R
0
f(x)
x z
0
¥
å
n=0

z z
0
x z
0

n
dx
=
1
2pi
I
jxz
0
j=R
0
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
dx
Next we want to see whether we can switch the integral sign
I
jxz
0
j=R
0
and the in?nite
summation
¥
å
n=0
. For this we need to show uniform convergence of the series below.
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
.
We use Weiestrass’s M-test: for any x on the circlefjx z
0
j = R
0
g, we have:




f(x)(z z
0
)
n
(x z
0
)
n+1









(z z
0
)
n
(x z
0
)
n+1




sup
jxz
0
j=R
0
j f(x)j
| {z }
=:C
R
0
=
C
R
0
R
0

jz z
0
j
R
0

n
Sincejz z
0
j< R
0
, the series
¥
å
n=0
C
R
0
R
0

jz z
0
j
R
0

n
converges. Note that the above series does not depend on x (the integration variable).
Hence by Weiestrass’s M-test, the series
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
converges uniformly on
the circlefjx z
0
j = R
0
g, thus allowing the switch between the integral sign and the
summation sign in (4.1):
f(z)=
1
2pi
¥
å
n=0
I
jxz
0
j=R
0
f(x)(z z
0
)
n
(x z
0
)
n+1
dx
=
¥
å
n=0
1
2pi

I
jxz
0
j=R
0
f(x)
(x z
0
)
n+1
dx

(z z
0
)
n
=
¥
å
n=0
f
(n)
(z
0
)
n!
(z z
0
)
n
.
In the last step we have used the higher order Cauchy’s integral formula. 
Example 4.1. The function f(z)= sin z is an entire function. By straight-forward
computations, its derivatives are given by:
f
0
(z)= cos z f
00
(z)= sin z
f
(3)
(z)= cos z f
(4)
(z)= sin z
.
.
.
.
.
.
Inductively, it is easy to deduce that f
(2k+1)
(0)=(1)
k
, and f
(2k)
(0)= 0 for any
integer k 0. Hence, the Taylor series of f about 0 is given by:
f(z)=
¥
å
k=0
f
(2k+1)
(0)
(2k+ 1)!
z
2k+1
=
¥
å
k=0
(1)
k
(2k+ 1)!
z
2k+1
= z
z
3
3!
+
z
5
5!

z
7
7!
+ . . .
This series converges to sin z for any z2C, because sin z is entire (i.e. holomorphic
on every ball B
R
(0)).
Example 4.2. Consider the function f(z) = Log(z) which is holomorphic on
W :=Cnfx+ 0i : x 0g. Note that we can only apply Theorem 4.1 if the ball
B
R
(z
0
) is contained insideW.
Let’s take z
0
= 1 as an example.
f
0
(z)=
1
z
f
0
(1)= 1
f
00
(z)=
1
z
2
f
00
(1)=1
f
(3)
(z)=
2
z
3
f
(3)
(1)= 2
f
(4)
(z)=
2 3
z
4
f
(4)
(1)=2 3
.
.
.
.
.
.
Inductively, we deduce that f
(n)
(1)=(1)
n1

(n 1)! for n 1.
Therefore, the Taylor series for f about 1 is given by:
Log(z)= Log(1)+
¥
å
n=1
(1)
n1

(n 1)!
n!
(z 1)
n
=
¥
å
n=1
(1)
n1
n
(z 1)
n
=(z 1)
1
2
(z 1)
2
+
1
3
(z 1)
3

1
4
(z 1)
4
+ . . .
Since f is holomorphic on B
1
(1) (but not on any larger ball centered at 1), the
above Taylor series converges to Log(z) on B
1
(1).
Example 4.3. The Taylor series for some composite functions, such as e
z
2
, can
be derived by substitution instead of deducing the general n-th derivative of the
function. For example:
e
z
= 1+ z+
z
2
2!
+
z
3
3!
+
z
4
4!
+ . . .
e
z
2
= 1+ z
2
+
(z
2
)
2
2!
+
(z
2
)
3
3!
+
(z
2
)
4
4!
+ . . .
= 1+ z
2
+
z
4
2!
+
z
6
3!
+
z
8
4!
+ . . .
Since the series for e
z
converges for any z2C, the series for e
z
2
converges for any
z2C as well.
Similarly, by replacing z by 1 z in the Taylor series for Log(z), we get:
Log(1 z)=z
1
2
z
2

1
3
z
3

1
4
z
4
 . . .
The series for Log(z) about 1 converges whenjz 1j< 1, and so the above series
for Log(1 z) converges whenj(1 z) 1j< 1, i.e.jzj< 1.
Apart from using Theorem 4.1 to ?nd the Taylor series of a given holomorphic
function, we can also make use of the geometric series formula directly:
1
1 w
= 1+ w+ w
2
+ . . . wherejwj< 1.
This method is particularly useful for functions whose n-th derivatives are tedious to
compute.
Page 5


Taylor and Laurent Series
4.1. Taylor Series
4.1.1. Taylor Series for Holomorphic Functions. In Real Analysis, the Taylor
series of a given function f :R!R is given by:
f(x
0
)+ f
0
(x
0
)(x x
0
)+
f
00
(x
0
)
2!
(x x
0
)
2
+
f
000
(x
0
)
3!
(x x
0
)
3
+ . . .
We have examined some convergence issues and applications of Taylor series in MATH
2033/2043. We also learned that even if the function f is in?nitely differentiable
everywhere onR, its Taylor series may not converge to that function. In contrast,
there is no such an issue in Complex Analysis: as long as the function f :C!C is
holomorphic on an open ball B
d
(z
0
), we can show the Taylor series of f :
f(z
0
)+ f
0
(z
0
)(z z
0
)+
f
00
(z
0
)
2!
(z z
0
)
2
+
f
000
(z
0
)
3!
(z z
0
)
3
+ . . .
converges pointwise to f(z) on B
d
(z
0
), and uniformly on any smaller ball. As we shall
see, it thanks to Cauchy’s integral formula. Moreover, the proof of Taylor Theorem
in Complex Analysis is also much easier than that in Real Analysis, again thanks to
Cauchy’s integral formula.
In this chapter, it is more convenient to re-label the variables in the Cauchy’s
integral formula:
f
(n)
(a)=
n!
2pi
I
g
f(z)
(za)
n+1
dz ! f
(n)
(z)=
n!
2pi
I
g
f(x)
(x z)
n+1
dx.
For the re-labelled Cauchy’s integral formula, we require the point z to be enclosed by
the simple closed curve g.
Theorem 4.1 (Taylor Theorem for Holomorphic Functions). Given a complex-valued
function f which is holomorphic on an open ball B
R
(z
0
), the series:
¥
å
n=0
f
(n)
(z
0
)
n!
(z z
0
)
n
converges (pointwise) to f(z) for any z2 B
R
(z
0
).
Proof. Given any z2 B
R
(z
0
), we let # > 0 be small enough so thatjz z
0
j < R#.
For simplicity, denote R
0
= R#.
By Cauchy’s integral formula, for any z2 B
R
0(z
0
), we have:
f(z)=
1
2pi
I
jxz
0
j=R
0
f(x)
x z
dx.
Then, the contourjz z
0
j = R
0
lies inside the open ball B
R
(z
0
). The key trick to prove
the Taylor Theorem is rewriting
1
x z
as a geometric series. Recall that:
1
1 w
= 1+ w+ w
2
+ . . . wheneverjwj< 1.
We ?rst rewrite
1
x z
into this form:
1
x z
=
1
(x z
0
)(z z
0
)
=
1
x z
0


1
1
zz
0
xz
0
=
1
x z
0
¥
å
n=0

z z
0
x z
0

n
Here we have used the fact that



zz
0
xz
0


< 1. See the diagram below. The yellow ball is
B
R
(z
0
), and the red circle isjx z
0
j = R
0
.
z
0
x
z
jx z
0
j = R
0
jz z
0
j< R
0
Then, whenever z2 B
R
0(z
0
), the function f(z) can be expressed as:
f(z)=
1
2pi
I
jxz
0
j=R
0
f(x)

1
x z
dx (4.1)
=
1
2pi
I
jxz
0
j=R
0
f(x)
x z
0
¥
å
n=0

z z
0
x z
0

n
dx
=
1
2pi
I
jxz
0
j=R
0
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
dx
Next we want to see whether we can switch the integral sign
I
jxz
0
j=R
0
and the in?nite
summation
¥
å
n=0
. For this we need to show uniform convergence of the series below.
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
.
We use Weiestrass’s M-test: for any x on the circlefjx z
0
j = R
0
g, we have:




f(x)(z z
0
)
n
(x z
0
)
n+1









(z z
0
)
n
(x z
0
)
n+1




sup
jxz
0
j=R
0
j f(x)j
| {z }
=:C
R
0
=
C
R
0
R
0

jz z
0
j
R
0

n
Sincejz z
0
j< R
0
, the series
¥
å
n=0
C
R
0
R
0

jz z
0
j
R
0

n
converges. Note that the above series does not depend on x (the integration variable).
Hence by Weiestrass’s M-test, the series
¥
å
n=0
f(x)(z z
0
)
n
(x z
0
)
n+1
converges uniformly on
the circlefjx z
0
j = R
0
g, thus allowing the switch between the integral sign and the
summation sign in (4.1):
f(z)=
1
2pi
¥
å
n=0
I
jxz
0
j=R
0
f(x)(z z
0
)
n
(x z
0
)
n+1
dx
=
¥
å
n=0
1
2pi

I
jxz
0
j=R
0
f(x)
(x z
0
)
n+1
dx

(z z
0
)
n
=
¥
å
n=0
f
(n)
(z
0
)
n!
(z z
0
)
n
.
In the last step we have used the higher order Cauchy’s integral formula. 
Example 4.1. The function f(z)= sin z is an entire function. By straight-forward
computations, its derivatives are given by:
f
0
(z)= cos z f
00
(z)= sin z
f
(3)
(z)= cos z f
(4)
(z)= sin z
.
.
.
.
.
.
Inductively, it is easy to deduce that f
(2k+1)
(0)=(1)
k
, and f
(2k)
(0)= 0 for any
integer k 0. Hence, the Taylor series of f about 0 is given by:
f(z)=
¥
å
k=0
f
(2k+1)
(0)
(2k+ 1)!
z
2k+1
=
¥
å
k=0
(1)
k
(2k+ 1)!
z
2k+1
= z
z
3
3!
+
z
5
5!

z
7
7!
+ . . .
This series converges to sin z for any z2C, because sin z is entire (i.e. holomorphic
on every ball B
R
(0)).
Example 4.2. Consider the function f(z) = Log(z) which is holomorphic on
W :=Cnfx+ 0i : x 0g. Note that we can only apply Theorem 4.1 if the ball
B
R
(z
0
) is contained insideW.
Let’s take z
0
= 1 as an example.
f
0
(z)=
1
z
f
0
(1)= 1
f
00
(z)=
1
z
2
f
00
(1)=1
f
(3)
(z)=
2
z
3
f
(3)
(1)= 2
f
(4)
(z)=
2 3
z
4
f
(4)
(1)=2 3
.
.
.
.
.
.
Inductively, we deduce that f
(n)
(1)=(1)
n1

(n 1)! for n 1.
Therefore, the Taylor series for f about 1 is given by:
Log(z)= Log(1)+
¥
å
n=1
(1)
n1

(n 1)!
n!
(z 1)
n
=
¥
å
n=1
(1)
n1
n
(z 1)
n
=(z 1)
1
2
(z 1)
2
+
1
3
(z 1)
3

1
4
(z 1)
4
+ . . .
Since f is holomorphic on B
1
(1) (but not on any larger ball centered at 1), the
above Taylor series converges to Log(z) on B
1
(1).
Example 4.3. The Taylor series for some composite functions, such as e
z
2
, can
be derived by substitution instead of deducing the general n-th derivative of the
function. For example:
e
z
= 1+ z+
z
2
2!
+
z
3
3!
+
z
4
4!
+ . . .
e
z
2
= 1+ z
2
+
(z
2
)
2
2!
+
(z
2
)
3
3!
+
(z
2
)
4
4!
+ . . .
= 1+ z
2
+
z
4
2!
+
z
6
3!
+
z
8
4!
+ . . .
Since the series for e
z
converges for any z2C, the series for e
z
2
converges for any
z2C as well.
Similarly, by replacing z by 1 z in the Taylor series for Log(z), we get:
Log(1 z)=z
1
2
z
2

1
3
z
3

1
4
z
4
 . . .
The series for Log(z) about 1 converges whenjz 1j< 1, and so the above series
for Log(1 z) converges whenj(1 z) 1j< 1, i.e.jzj< 1.
Apart from using Theorem 4.1 to ?nd the Taylor series of a given holomorphic
function, we can also make use of the geometric series formula directly:
1
1 w
= 1+ w+ w
2
+ . . . wherejwj< 1.
This method is particularly useful for functions whose n-th derivatives are tedious to
compute.
Example 4.4. Consider the function:
f(z)=
z 2
(z+ 2)(z+ 3)
.
We are going to derive its Taylor series about 0. First, we do partial fractions on
the function:
f(z)=
5
z+ 3

4
z+ 2
.
Then, we try to rewrite each term above in the form of
a
1w
. Note that:
5
z+ 3
=
5
3


1
z
3
+ 1
=
5
3


1
1


z
3


=
5
3
¥
å
n=0


z
3

n
=
¥
å
n=0
(1)
n
5
3
n+1
z
n
(wherejzj< 3)
4
z+ 2
=
4
2


1
z
2
+ 1
=
2
1


z
2


= 2
¥
å
n=0


z
2

n
(wherejzj< 2)
=
¥
å
n=0
(1)
n
2
n1
z
n
.
Hence, forjzj< 2, we have:
f(z)=
¥
å
n=0
(1)
n
5
3
n+1
z
n

¥
å
n=0
(1)
n
2
n1
z
n
=
¥
å
n=0
(1)
n

5
3
n+1

1
2
n1

z
n
.
To derive the Taylor series of f about other center (say 1), we can express
5
z+3
and
4
z+2
into:
5
z+ 3
=
5
(z 1)+ 4
=
5
4


1
1


z1
4

=
¥
å
n=0
5
4


z 1
4

n
(wherejz 1j< 4)
=
¥
å
n=0
(1)
n
5
4
n+1
(z 1)
n
4
z+ 2
=
4
(z 1)+ 3
=
4
3


1
1


z1
3

=
¥
å
n=0
4
3


z 1
3

n
(wherejz 1j< 3)
=
¥
å
n=0
(1)
n
4
3
n+1
(z 1)
n
.
Therefore, onjz 1j< 3, we have:
f(z)=
¥
å
n=0
(1)
n

5
4
n+1

4
3
n+1

(z 1)
n
.