# Essential Singularity

## 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

## 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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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:

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

therefore has an essential singularity at z=z0.

I'm not supposed to know that theorem yet -- is there a way to prove it using only Casaroti-Weierstrass?

$$e^{1/z}$$

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

$$e^{1/z}=\sum_{n=0}^{\infty} \frac{1}{z^n n!}$$

and the same would be true for:

$$e^{1/z+1/z^2}$$

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:

$$e^{1/z+1/z^2+\cdots+1/z^n}$$

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

$$f(z)=\frac{\phi(z)}{z^m}$$

with $\phi(z)$ analytic and not zero at zero so:

$$\phi(z)=\sum_{k=0}^{\infty} a_k z^k$$

then:

$$e^{f(z)}=\exp(\frac{\sum_{k=0}^{\infty} a_k z^k}{z^m})$$

$$=\exp(h(z)) \exp(\sum_{j=0}^{N}\frac{c_j}{z^j})$$

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