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Let

$l(\xi)=\xi_n^k + $ lower order terms in $\xi_n$

be a polynomial in $\mathbb{C}^n$, $\xi'=(\xi_1,\ldots,\xi_{n-1})$, and $l_{\xi'}(z)=l(i\xi',iz)$. Further, let $\lambda_1(\xi'),\ldots,\lambda_k(\xi')$ denote to the zeros of $l$.

Now, why is the mapping $\xi'\mapsto \lambda_i(\xi')$ continuous for all $i=1,\ldots,k$?

edit: This is part of a proof of the Malgrange-Ehrenpreis theorem. There it vaguely refers to Rouché's theorem. If it applies to the above-mentioned problem I still don't see how.

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    if $l(\xi)$ depends only on $\xi_n$, then how can $\lambda_i(\xi') = \lambda_i(\xi_1, ...,\xi_{n-1})$ denote the zeroes of $l$?2015-04-21

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