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Essential Singularity

  • Thread starter murmillo
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  • #1
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Homework Statement


Show that if z0 is an isolated singularity of f(z) that is not removable, then z0 is an essential singularity for ef(z).


Homework Equations


z0 is a pole of f(z) of order N iff f(z) = g(z)/(z-z0)^N, where g is analytic at z0 and g(z0) is not 0, iff 1/f(z) is analytic at z0 and has a zero of order N, iff |f(z)| → ∞ as z → z0.

Casorati-Weierstrass theorem

The Attempt at a Solution

Homework Statement


I know what to do if z is an essential singularity. But I'm having trouble if z is a pole. It seems that what I must do is find a sequence of points going to z0 such that f(z) goes to infinity, and also a sequence of points going to z0 such that f(z) approaches some complex number. I don't know how to do that.
 
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Answers and Replies

  • #2
1,796
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You could rely on Picard's First Theorem: Any entire function that's not a polynomial has an essential singularity at infinity. So Exp(z) is an entire function and:

[tex]Exp[\frac{A(z)}{(z-z_0)^n}] [/tex]

therefore has an essential singularity at z=z0.
 
  • #3
118
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I'm not supposed to know that theorem yet -- is there a way to prove it using only Casaroti-Weierstrass?
 
  • #4
1,796
53
Ok, how about this then: You'd agree that the function:

[tex]e^{1/z}[/tex]

has an essential singularity at zero right since the Laurent series is:

[tex]e^{1/z}=\sum_{n=0}^{\infty} \frac{1}{z^n n!}[/tex]

and the same would be true for:

[tex]e^{1/z+1/z^2}[/tex]

since you could multiply the individual sums for e^(1/z) and e^(1/z^2) and still end up with a series with an infinite number of terms with z in the denominator. And we could keep doing that to show that:

[tex]e^{1/z+1/z^2+\cdots+1/z^n}[/tex]

also has an essential singularity at zero. Ok then, so for simplicity assume f(z) has a pole at zero so we can write:

[tex]f(z)=\frac{\phi(z)}{z^m}[/tex]

with [itex]\phi(z)[/itex] analytic and not zero at zero so:

[tex]\phi(z)=\sum_{k=0}^{\infty} a_k z^k[/tex]

then:

[tex]e^{f(z)}=\exp(\frac{\sum_{k=0}^{\infty} a_k z^k}{z^m})[/tex]

[tex]=\exp(h(z)) \exp(\sum_{j=0}^{N}\frac{c_j}{z^j})[/tex]

with h(z) now analytic and e^(h(z)) non-zero. What then can you say about the other part of that expression?
 

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