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G c |C; f : G -> |C; b:-|C
Suppose that function f has a pole at point b.
Can it also have a removable singularity at b?
NO.
By virtue of our supposition:
(1) a doesn't belong to G
(2) there exists B = {z:-|C : 0 < |z-a| < r} c G such that f is diffable on B
(3) lim(z->b) |f(z)| = oo
G c |C; f : G -> |C; b:-|C
What does it mean that function f has an essential isolated singularity at point b?
(1) f has an isolated singularity at b
(2) the singularity is not removable
(3) the singularity is not a pole
Consider the complex function f(z)=exp(1/z).
What kind of isolated singularity does it have at 0?
essential
(1) lim(n->oo) |f(1/n)| = oo
(2) lim(n->oo) |f(i/n)| = 1 < oo
(1) not removable
(2) not a pole
Let m be a positive integer.
Consider the complex function f(z)=z^m * sin(1/z).
What kind of isolated singularity does it have at 0?
essential
(1) lim(n->oo) |f(i/n)| = oo
(2) lim(n->oo) |f(1/n)| = 0
(1) not removable
(2) not a pole
Consider the complex function f(z) = sin(z)/z.
What kind of isolated singularity does it have at 0?
removable
====================
You can prove your answer by referring to the definition only.
This is also an opportunity to test the unproven theorem. (MRW 2001.02.27)
THEOREM: G c |C; f : G -> |C; b:-|C
thesis: If function f has an isolated singularity at poin
Consider the complex function f(z) = ( cos(z) - 1 ) / z.
What kind of isolated singularity does it have at 0?
removable
====================
You can prove your answer by referring to the definition only.
This is an opportunity to test the unproven theorem. (MRW 2001.02.04)
THEOREM: G c |C; f : G -> |C; b:-|C
thesis: If function f has an isolated singularity at poin
lim(x->0,y->0) (x-sin(y))*y^2 / (y^2 + sin(x-y)*sin(x-y) ) = ???
= 0
hint: y^2 <= y^2 + sin(x-y)*sin(x-y)
lim(x->0) x*sin(x) / ( x^2 + sin(x)*sin(x) ) = ???
= 1/2
hint: use the squeeze theorem
Consider the complex function f(z) = (z+1)/z.
Does it have a removable isolated singularity at 0 ?
Justify your answer in two different ways.
NO. It actually has a pole. Notice that lim(z->0) |f(z)| = oo.
---------------------------------
(1) lim(z->0) z*f(z) = 1
hence not (2) lim(z->0) (z-0)*f(z) = 0.
And (2) is a necessary condition for a removable isolated singularity.
(page 103, Conway, "Functions of One Complex Variable", cha
w:-|C
2*sin^2(w) = ???
2*sin^2(w) = 1 - cos(2*w)

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