Chapter 10: Taylors theorem

Question 1

THM 10.1 (Taylor’s Theorem)

Answer 1

Let the function f be analytic in the disc
∆ = {z ∈ C : |z − z0| < r}, where r > 0 . Then f(z) has a Taylor expansion about z0
valid on ∆.

For all z ∈ ∆,

f(z) = f(z_0) +
∑from n=1 to ∞ of [f^{n}(z_0)/n!] (z-z_0)^n

Question 2

EXAMPLE:Find the Taylor series expansion of 1/(1−z^2) about z = 0. Where is this expansion valid?

Answer 2

1/ (1-t) expansion gives

1/(1−z^2) = 1 +z^2 + z^4 + z^6 +…
= ∑from n=1 to ∞ of z^{2n}
for |z| less than 1

This is the Taylor series for 1/(1−z^2) valid in the disc D = {z ∈ C : |z| < 1}.

Question 3

1/ (1-t)

Answer 3

1/ (1-t)= 1 + t + t^2 + t^3 + · · ·

|t| < 1

Question 4

Theorem 10.2 (Uniqueness Theorem for Taylor series expansions)

Answer 4

f(z) = b_0 + b_1 (z-z_0) + b_2 (z-z_0)^2 + …

Suppose that the
function f is analytic in the disc ∆ = {z ∈ C : |z − z0| < r}, where r > 0. Suppose
also that

∑from n=1 to ∞ of b_n(z-z_0)^n on ∆.

Then
b_n = f^{n} (z_0) /n!
n=0,1,2,..

ie its the Taylor series for f(z) on ∆

Question 5

EXAMPLE: Find the Taylor series of 1/(z − 1)

about z = 2.

Answer 5

1/(z − 1) is analytic on C{1} and so analytic on the discD = {z ∈ C : |z − 2| < 1}. This is the largest disc with centre 2 in which f is
analytic.

let w=z-2

then for |z-2| less than 1 ie |w| less than 1

1/(z − 1) = 1/(w+1) = ∑from n=1 to ∞ of (-1)^n w^n

so the Taylor series about z = 2, in powers of (z − 2), is

1/(z − 1) = ∑from n=1 to ∞ of (-1)^n (z-2)^n

valid on ∆ = {z : |z − 2| < 1}

Question 6

about z_0

Answer 6

In general, if the centre of the expansion is z0 6= 0, put w = z−z0 and expand
in powers of w, then replace w by (z − z0), to obtain a power series in powers
of (z − z0).

Question 7

Def inition 10.3 ZERO

Answer 7

Suppose that f is analytic on a region D and f(w) = 0 for some w ∈ D.
Then w is a zero of f.

Question 8

DEF zero of order k

Answer 8

Suppose that the function f is analytic in a region D and w ∈ D.
If f(w) = f'(w) = · · · = f^(k−1)(w) = 0 and f^(k) (w) ≠ 0 (so we can take out a factor
(z − w)^k), we say that f has a zero of order k at w.

Question 9

THM 10.5 zero order k

Answer 9

Let k be a positive integer. The function f has a zero of order k at w if
and only if
f(z) = (z − w)^k g(z)
in some neighbourhood U of w, where the function g is analytic and non-zero on U

Question 10

COROLLARY 10.6 zeros order and products of functs

Answer 10

Suppose that
(i) the function f has a zero of order m at the point w;

(ii) the function g has a zero of order n at the point w.

Write h(z) = f(z)g(z). Then h has a zero of order m + n at w.
i.e. fg has a zero of order m + n at the point w.

Question 11

EXAMPLE:

Show that sin z has a simple zero at z = 0.

Answer 11

Let f(z) = sin z Then, f is analytic in C and, for all integers n,
f(nπ) = sin(nπ) = 0 and f’(nπ) = (cos(nπ)) = (−1)^n ≠ 0.
Thus sin z has a simple
zero at z = 0. It also has simple zeros at the points
± π, ± 2π, ± 3π, · · · .

Question 12

EXAMPLE:

Find the order of the zero of 1 − cos z at z = 0.

Answer 12

Let g(z) = 1 − cos z. Then g is analytic in C and
g(0) = 0 , g'(0) = sin 0 = 0 , g''(0) = cos 0 = 1 ≠ 0.
Hence g has a zero of order 2 at z = 0.

Question 13

EXAMPLE:

Show that z sin(z
2
) has a zero of order 3 at the origin.

Answer 13

Let h(z) = sin(z^2)
Then h is analytic on C and
h(0) = 0,
h’(0) =(2z cos(z^2))_z=0 = 0, h’‘(0) = [2 cos(z^2) − 4z^2sin(z^2)]z=0 = 1 ≠ 0.

Thus sin(z^2) has a zero of order 2 at the origin. Since z is analytic in C and has a
zero of order 1 at the origin, we see, from corollary 10.6, that z sin(z^2) has a zero of
order 3 at the origin

Question 14

simple zero and why zeros

Answer 14

Thus if the function f has a zero of order k at w, then we can express f(z) in the form
f(z) = (z−w)^k g(z) where g is analytic in some neighbourhood of w and g(w) = f^(k)(w)/k! ≠ 0.
If k = 1, we say that the zero is a simple zero.