MHB What does zero at infinity mean?

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The discussion centers on the concept of zeros at infinity in relation to the Laplace transform of signals. It clarifies that a zero at infinity occurs when a function approaches zero as its variable approaches infinity, exemplified by the function \( f(z) = z^{-2} \). The transformation of \( f(z) \) into \( g(z) \) allows for counting the multiplicity of zeros at zero, which corresponds to the zeros at infinity for the original function. The conclusion drawn is that \( f(z) \) has two zeros at infinity based on this analysis. Understanding zeros at infinity is crucial for analyzing the behavior of functions in the context of complex variables and Laplace transforms.
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For the following algebraic expressions for the Laplace transform of a signal, determine the number of zeros located in the finite \(s-\)plane and the number of zeros located at infinity.

Does that mean pole since at a pole the plot will diverge to \(\infty\)? Otherwise, what would be a zero at infinity?
 
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Consider the complex-valued function of a complex variable, $f(z) = z^{-2}$. We notice that as $|z|\to \infty$ that $f(|z|) \to 0$ so we say $f(z)$ is zero at infinity. I think when someone asks for "number of zeros at infinity" they are asking for the multiplicity of $\infty$ as a zero.

In general, if $f(z)$ sends $\infty$ to zero then,
$$ \lim_{z\to 0} f\left( \frac{1}{z} \right) = 0 $$
We can define the function,
$$ g(z) = \left\{ \begin{array}{ccc} f(z^{-1}) & \text{ if }& z\not = 0 \\ 0 & \text{ if }& z = 0 \end{array} \right. $$
Now count the multiplicity of $0$ for $g(z)$ at $0$.

In the example given above $f(z)=z^{-2}$ then $g(z) = z^2$ so $g(z)$ has multiplicity two zero. Thus, $f(z)$ has two zeros at infinity.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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