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Going through Bax and Newman's Complex Analysis book, and found this question:

Let $P$ be a nonconstant polynomial in $z$. Show that $P(z) \rightarrow \infty$ as $z \rightarrow \infty$.

The solution in the book cryptically suggests just considering $P(z) = z^n(a_n + \frac{a_{n-1}}{z} + ... + \frac{a_0}{z^n})$. But how can we derive from this that for a given $z$ s.t. $|z| > M$, $|P(z)| > N$? Intuitively it makes sense, but one can't apply the triangle theorem here to get $|P(z)|$ in terms of $|z^n|$. I find it hard to use the fact that $|Re(z)| < |z|$ here since it's hard to really pinpoint the real component of a polynomial in $z$. How can one show that $|P(z)| > N$ in this case?

O M
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SalmonKiller
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  • If $|z|$ is large enough then the stuff in brackets is bounded below in absolute value by $|a_n|/2$ which is a positive number. – peek-a-boo Nov 14 '23 at 07:33
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    $a_n + \frac{a_{n-1}}{z} + ... + \frac{a_0}{z^n}\to a_n$. – geetha290krm Nov 14 '23 at 07:34
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    For what it's worth, the conjecture should be restated as $~|z| \to \infty \implies |P(z)| \to \infty.$ – user2661923 Nov 14 '23 at 07:35
  • @geetha290krm how would you show that formally? It makes sense that as $|z| \rightarrow \infty$, $\frac{a_0}{|z|} \rightarrow 0$, but how would it follow that $|\frac{a_0}{z} + \frac{a_1}{z^n}| \rightarrow 0$? – SalmonKiller Nov 14 '23 at 07:43
  • $|a_n + \frac{a_{n-1}}{z} + ... + \frac{a_0}{z^n}|\ge |a_n| -\epsilon (|a_{n-1}|+...+|a_0|)$ if $|z|>\frac 1{\epsilon}$. – geetha290krm Nov 14 '23 at 07:45
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    https://math.stackexchange.com/q/1162261/42969, https://math.stackexchange.com/q/39567/42969, https://math.stackexchange.com/q/3557923/42969 – Martin R Nov 14 '23 at 07:51

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