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- #1

Does that mean pole since at a pole the plot will diverge to \(\infty\)? Otherwise, what would be a zero at infinity?

- Thread starter dwsmith
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- Thread starter
- #1

Does that mean pole since at a pole the plot will diverge to \(\infty\)? Otherwise, what would be a zero at infinity?

- Jan 26, 2012

- 236

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.