What does zero at infinity mean?

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SUMMARY

The discussion clarifies the concept of zeros at infinity in the context of Laplace transforms and complex functions. Specifically, it establishes that for the function \(f(z) = z^{-2}\), as \(z\) approaches infinity, the function approaches zero, indicating that it has two zeros at infinity. The transformation of \(f(z)\) into \(g(z) = z^2\) demonstrates that \(g(z)\) has a multiplicity of two at zero, confirming the original function's zeros at infinity. This analysis is crucial for understanding the behavior of functions in the complex plane.

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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.
 

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