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Assume $p$ is a non constant polynomial of degree $n$. Prove that the set $\{z:|(p(z))| \lt 1\}$ is a bounded open set with at-most $n$ connected components. Give example to show number of components can be less than $n$.


EDIT:Thanks,I meant connected components.

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up vote 2 down vote accepted


  • For boundedness: Show that $|p(z)| \to \infty$ as $|z|\to \infty$
  • For openness: The preimage $p^{-1}(A)$ of an open set $A$ under a continuous function $p$ is again open. Polynomials are continuous, so just write your set as a preimage.

  • For the connected components: Recall fundamental theorem of algebra. What may these components have to do with polynomial roots? How many of them can a polynomial have? The example for fewer components is straightforward, just think of it geometrically (draw a curve).

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Thanks for reply.I can prove boundedness and openness,main deal is connected components,Cannot answer your question 2,please elaborate.. – Random Sep 8 '12 at 10:19

For bounding the number of connected components, you can show that every connected component contains at least one root of your polynomial. To obtain that, you can apply the minimum modulus principle to the connected component, just like in a proof of the fundamental theorem of algebra.

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