NPTEL web course
on
Complex Analysis
A. Swaminathan
I.I.T. Roorkee, India
and
V.K. Katiyar
I.I.T. Roorkee, India
A.Swaminathan and V.K.Katiyar (NPTEL)
Complex Analysis
1 / 17
Complex Analysis
Module: 5: Consequences of Complex Integration
Lecture: 2: Taylor series
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Complex Analysis
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Consequences of complex integration
Taylor Series
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Complex Analysis
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Taylor Series
In this part, we find certain series expansions that are consequences
of Cauchy Integral Formula.
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Complex Analysis
4 / 17
Taylor Series
Taylor theorem
Theorem
Let f (z) be analytic at all points within a circle C0 with center z0 and
radius ρ0 . Then for every point z within C0 , we have
f 00 (z0 )
fn
(z − z0 )2 + . . . + (z − z0 )n + . . .
2!
n!
∞
∞
n
n
X
X
(z − z0 ) (n)
(z − z0 ) (n)
= f (z0 ) +
f (z0 ) =
f (z0 ),
n!
n!
f (z) = f (z0 ) + f 0 (z0 )(z − z0 ) +
n=1
n=0
where the power series converges to f (z) on the disc |z − z0 | < ρ0 .
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Complex Analysis
5 / 17
Taylor Series
The representation of f (z) as an infinite series
∞
X
(z − z0 )n (n)
f (z0 ) is known as Taylor Series.
f (z) =
n!
n=0
Using the Taylor series expansion, we can write
Z
1
f (z)
f (n) (z0 ) =
dz.
2πi c (z − z0 )n+1
We prove this result using Cauchy Integral formula.
When z0 = 0, this reduces to
f (z) = f (0) +
∞
X
zn
n=1
n!
f (n) (0)
which is known as Maclaurin’s series.
Taylor’s series at origin is called Maclaurin’s series.
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Complex Analysis
6 / 17
Taylor Series
For the validity of the expansion as a Taylor series, it is essential
that f (z) be analytic at all points inside the circle C0 for then the
convergence of Taylor series for f (z) is assured.
Hence the greatest radius of C0 is the distance from the point z0 to
the singularity of f (z) which is nearest to z0 , since we require the
function to be analytic at all points within C0 .
If f (z) is analytic in a region Ω, containing the point ’a’ such that
f (a) and all its derivatives f n (a) vanishes, then f(z)=0 in Ω.
If f is analytic at z0 , the Taylor series for f 0 around z0 can be
obtained by termwise differentiation of the Taylor series for f
around z0 and converges in the same disk as the series for f .
A.Swaminathan and V.K.Katiyar (NPTEL)
Complex Analysis
7 / 17
Taylor Series
Let f and g be analytic functions with Taylor series
∞
∞
X
X
aj (z − z0 )j and g(z) =
bj (z − z0 )j around the point
f (z) =
j=0
j=0
z0 , then
(i) the Taylor series for cf(z), c a constant, is
∞
X
caj (z − z0 )j ;
j=0
(ii) the Taylor series for f (z) ± g(z) is
∞
X
(aj ± bj )(z − z0 )j .
j=0
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Complex Analysis
8 / 17
Taylor Series
The Cauchy product of two Taylor series f (z) =
∞
X
aj (z − z0 )j
j=0
and g(z) =
∞
X
bj (z − z0 )j is defined to be the series
j=0
h(z) =
∞
X
cj (z − z0 )j , where cj =
j=0
j
X
aj−i bi .
i=0
Let f and g be analytic functions with Taylor series
∞
∞
X
X
f (z) =
aj (z − z0 )j and g(z) =
bj (z − z0 )j around the point
j=0
j=0
z0 . Then the Taylor series for the product fg around z0 is given by
the Cauchy product of these two series.
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Complex Analysis
9 / 17
Taylor Series
Comparison with real functions
An analytic function is infinitely differentiable and always has a
Taylor series expansion with a non-zero radius of convergence.
A real function f can also be infinitely differentiable. But it is
sometimes impossible to expand it in a power series.
An example of such a function is
(
2
e−1/x ,
f (x) =
0,
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Complex Analysis
x=
6 0
x = 0.
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Taylor Series
Comparison with real functions
If a real valued function f (x) defined on the whole of real line is
extended as complex valued function f (z) analytically, it may not
be analytic in the entire complex plane.
For example, ex is real analytic function for all real x and its
complex extension ez is also an analytic (entire) function.
Whereas, 1/(1 + x 2 ) is a real analytic function for all real x. But its
complex extension 1/(1 + z 2 ) ceases to be analytic at z = ±i
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Complex Analysis
11 / 17
Taylor Series
Example
The function f (z) = exp(z) is entire. Hence Macluaurin series
expansion is valid for all z. Since f (n) (0) = 1 we get
f (z) = exp(z) =
∞
X
zn
n=0
A.Swaminathan and V.K.Katiyar (NPTEL)
Complex Analysis
n!
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Taylor Series
Example
The function f (z) = sin z is entire. Hence Macluaurin series expansion
is valid for all z. Since f (2n) (0) = 0 and f (2n−1) (0) = (−1)n , we get
f (z) = sin z =
∞
X
(−1)n
n=0
z 2n+1
(2n + 1)!
Differentiating both sides of the function gives
cos z =
∞
X
(−1)n
n=0
z 2n
(2n)!
Since sinh z = −i sin(iz), the Maclaurin series expansion for
sinh z and cosh z are also possible.
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Complex Analysis
13 / 17
Taylor Series
Example
Expand log(1 + z) in a Taylor Series about z=0 and determine the
region of convergence for the resulting series.
Solution:
Here f (z) = log(1 + z). We find f 0 (z), f 00 (z), f 000 (z) . . . f n (z) and their
values at the point 0.
Using the above values, we write the series as
f (z) = z −
z2 z3 z4
zn
+
+
+ . . . + (−1)n−1
+ ...
2
3
4
n
The point z = −1 is the singularity of log(1 + z) nearest the point
z=0. Thus the series converges for all values of z within the circle
|z| = 1.
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Complex Analysis
14 / 17
Taylor Series
Example
Expand
f (z) =
1
1−z
about the point z0 = 2i.
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Complex Analysis
15 / 17
Taylor Series
We illustrate the use of geometric series in deriving the Taylor series.
Solution
Write
1
1
=
1−z
1 − z + 2i − 2i
1
1
=
1 − 2i 1 − z−2i
1−2i
1
1
1
=
+
(z − 2i) +
(z − 2i)2 + · · ·
1 − 2i
(1 − 2i)2
(1 − 2i)3
The distance
√ from the center z0 = 2i to the nearest singularity
z = 1, is 5. Hence the circle of convergence is
√
|z − 2i| < 5
A.Swaminathan and V.K.Katiyar (NPTEL)
Complex Analysis
16 / 17
Taylor Series
Exercises
Expand in a Maclaurin series and also give the radius of
convergence of the function
f (z) = ze−z
2
Use Maclaurin series for ez to expand
f (z) = (z − 1)ez
about z0 = 1.
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Complex Analysis
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