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The theorem says:

"Suppose $z_0$ is an essential isolated singularity of $f(z)$. Then for every complex number $w_0$, there is a sequence $z_n\rightarrow z_0$ such that $f(z_n)\rightarrow w_0$."

The function $f(z)=e^{1/z}$ has an essential singularity at $z=0$. Can someone demonstrate the theorem up above by providing a sequence of complex numbers $z_n$ so that:

$$z_n\rightarrow 0 \qquad\text{and}\qquad f(z_n)\rightarrow 10$$

And perhaps a second example where:

$$z_n\rightarrow 0 \qquad\text{and}\qquad f(z_n)\rightarrow 1+i$$

Thanks.

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Let $\ln(10)$ be the real natural logarithm of 10. Then for $$z_n = \frac{1}{\ln(10) + 2\pi n i}$$ we have $z_n \to 0$ and $f(z_n) = 10$. You can do the same thing for $1 + i$ by replacing $\log(10)$ with any number such that $e^z = 1 + i$.

As a comment, by the Picard theorem, there is at most one complex number $w_0$ with the property that we cannot choose $z_n \to z_0$ with $f(z_n) = w_0$. In this case it is $0$.

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  • $\begingroup$ Very helpful. Thanks. $\endgroup$ – David Apr 21 '13 at 2:32

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