Complex Variables Limit Problem(s)


by wtmore
Tags: complex, limit, variables
wtmore
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#1
Feb26-12, 05:31 PM
P: 4
1. The problem statement, all variables and given/known data
a) [tex]\lim_{z\to 3i}\frac{z^2 + 9}{z - 3i}[/tex]
b) [tex]\lim_{z\to i}\frac{z^2 + i}{z^4 - 1}[/tex]


2. Relevant equations
????


3. The attempt at a solution
I'm assuming both of these are very, very similar, but I'm not quite sure how to solve them. I would like a method other than using ε and [itex]\delta[/itex].

If you simply plug in the limit, it's obviously indeterminate. Is there an easy method to solve these limits or is the only option to use ε and [itex]\delta[/itex]? I'm not sure how to start, any suggestions would be helpful. Thanks.
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Ansatz7
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#2
Feb26-12, 05:33 PM
P: 29
Try factoring the numerator and/or denominators. It's quite simple from there.
wtmore
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#3
Feb26-12, 05:42 PM
P: 4
Wow, can't believe I didn't realize that. It helped me solve a), which I ended up getting to be 6i, but b) cannot be factored (I don't think?).

If it were [tex]z^4 + 1[/tex] in the denominator then I could, but I'm pretty sure I cannot factor anything in that problem?

SteveL27
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#4
Feb26-12, 05:52 PM
P: 799

Complex Variables Limit Problem(s)


Quote Quote by wtmore View Post
Wow, can't believe I didn't realize that. It helped me solve a), which I ended up getting to be 6i, but b) cannot be factored (I don't think?).

If it were [tex]z^4 + 1[/tex] in the denominator then I could, but I'm pretty sure I cannot factor anything in that problem?
The denominator is a difference of squares. Then one of the factors has the same type of factorization as (a), namely the trick that a *sum* of squares can be factored with the use of an imaginary number.
wtmore
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#5
Feb26-12, 06:04 PM
P: 4
Quote Quote by SteveL27 View Post
The denominator is a difference of squares. Then one of the factors has the same type of factorization as (a), namely the trick that a *sum* of squares can be factored with the use of an imaginary number.
So I have:
[tex]\frac{z^2+i}{z^4-1}=\frac{z^2+i}{(z^2-1)(z^2+1)}=\frac{z^2+i}{(z-1)(z+1)(z-i)(z+i)}[/tex]
Am I missing something in the numerator?

EDIT: Would multiplying by the numerators conjugate be beneficial?
SteveL27
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#6
Feb26-12, 06:17 PM
P: 799
Quote Quote by wtmore View Post
So I have:
[tex]\frac{z^2+i}{z^4-1}=\frac{z^2+i}{(z^2-1)(z^2+1)}=\frac{z^2+i}{(z-1)(z+1)(z-i)(z+i)}[/tex]
Am I missing something in the numerator?

EDIT: Would multiplying by the numerators conjugate be beneficial?
Hmmm ... that didn't help. I'm stuck too now.
Dick
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#7
Feb26-12, 06:36 PM
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Quote Quote by SteveL27 View Post
Hmmm ... that didn't help. I'm stuck too now.
The second one isn't indeterminant.
wtmore
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#8
Feb26-12, 06:41 PM
P: 4
Quote Quote by Dick View Post
The second one isn't indeterminant.
Would the answer be [itex]\pm∞[/itex]?
SteveL27
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#9
Feb26-12, 06:44 PM
P: 799
Quote Quote by wtmore View Post
Would the answer be [itex]\pm∞[/itex]?
Just [itex]∞[/itex]. There's only one point at infinity in the extended complex numbers. If you think of the plane, you go to infinity when you go toward the edge of the plane in any direction. There's only one complex infinity, way out there beyond the edge of the plane.

A really nice visualization is to add a single point at infinity, and identify it with the "circumference" of the plane ... take the entire plane and fold it into a sphere, with the point at infinity at the north pole. It's called the Riemann sphere.

http://en.wikipedia.org/wiki/Riemann_sphere
Dick
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#10
Feb26-12, 06:45 PM
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Quote Quote by wtmore View Post
Would the answer be [itex]\pm∞[/itex]?
Not in the complex numbers. It's a pole. Saying "does not exist" is probably safe.


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