Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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 $0$.

share|improve this answer
    
Very helpful. Thanks. –  David Apr 21 '13 at 2:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.