Suppose that f, g € H(D(z, r)), 1 < m < n, f has a zero of orderz and g has a zero of order m at z. Show that f/g has a removablelarity at z.

Name: Suppose that f, g € H(D(z, r)), 1 < m < n, f has a zero of orderz and
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Description: Suppose that f, g € H(D(z, r)), 1 < m < n, f has a zero of orderz and g has a zero of order m at z. Show that f/g has a removablelarity at z.

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Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Suppose that f, g € H(D(z, r)), 1 < m < n, f has a zero of order
z and g has a zero of order m at z. Show that f/g has a removable
larity at z.

Expert Solution

Step 1

An isolated singularity $z_{0}$ of the function is called removable if the function can be defined at so that it become analytic at $z_{0}$ .

From the question f and g are analytic on an open set D and f has a zero of order n and g has a zero of order m at z .

let the function f and g are defined as:

$f (z) = z^{n} ϕ (z), ϕ (z) i s a n a l y t i c a n d n o n - z e r o i n t h e n e i g h b o u r h o o d o f z = 0 g (z) = z^{m} ψ (z) ψ (z) i s a n a l y t i c a n d n o n - z e r o i n t h e n e i g h b o u r h o o d o f z = 0$

Step 2

Now the function ^{$\frac{f}{g}$} is defined as:

$f (z) = z^{n} ϕ (z), g (z) = z^{m} ψ (z) \frac{f (z)}{g (z)} = \frac{z^{n} ϕ (z)}{z^{m} ψ (z)} = \frac{z^{n - m} ϕ (z)}{ψ (z)}, w h e r e n \geq m \geq 1$

Step 3

$S i n c e, ϕ (z) a n d ψ (z) i s a n a l y t i c a n d n o n - z e r o i n t h e n e i g h b o u r h o o d o f z = 0 \frac{f (z)}{g (z)} = \frac{z^{n - m} ϕ (z)}{ψ (z)}, w h e r e n \geq m \geq 1 n o w l e t t h e c a s e n = m, t h e n \frac{f (z)}{g (z)} = \frac{z^{n} ϕ (z)}{z^{n} ψ (z)} = ρ (z) \neq 0 i n t h e n e i g h b o u r h o o d o f z = 0 t h u s, \lim_{z \to 0} z \frac{f (z)}{g (z)} = \lim_{z \to 0} z ρ (z) = 0 i n d i c a t i n g t h a t z = 0 i s a r e m o v a b l e \sin g u l a r i t y$

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