Solved: "Let a0,a1,…,an−1,an complex numbers with an≠0. Iff(z)=|an+an−1z+⋯+a0zn|Exist lim|z|→∞f(z)?"

cleffavw8

Answered question

2022-03-23

Let

a_{0}, a_{1}, \dots, a_{n - 1}, a_{n}

complex numbers with

a_{n} \neq 0

. If

f (z) = | a_{n} + \frac{a_{n - 1}}{z} + \dots + \frac{a_{0}}{z^{n}} |

Exist

lim_{| z | \to \infty} f (z)

Gia Edwards

Beginner2022-03-24Added 12 answers

If we let

z \to \infty

, intuitively then we expect the terms containing z in the denominator to drop out leaving the limit as

| a_{n} |

. Let's see how our intuition holds out. We have

| | a_{n} + \frac{a_{n - 1}}{z} + \dots + \frac{a_{0}}{z^{n}} | - | a_{n} | | \leq | \frac{a_{n - 1}}{z} + \dots + \frac{a_{0}}{z^{n}} |

The above inequality follows from the reverse triangl inequality. Now apply the regular triangle inequality

| \frac{a_{n - 1}}{z} + \dots + \frac{a_{0}}{z^{n}} | \leq | \frac{a_{n - 1}}{z} | + \dots + | \frac{a_{0}}{z^{n}} |

If the proof is rather informal at this point you can just say that as

z \to \infty

the right hand side above tends to 0 and the limit holds. Otherwise, it wouldn't be too difficult to follow up with a formal

ϵ - δ

proof.

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