Question

If p(z) = a_0 + a_1z + ..... + a_{n-1}z^{n-1} + z^n(n \geq 1) , then show that there exists a real R > 0 such that 2^{-1}\lvert z \rvert^n \leq \lvert p(z) \rvert \leq 2 \lvert z \rvert^n for \lvert z \rvert \geq R

07 Feb 2024
Answer :
Word Count : 409

To prove this inequality, let's consider the expression \( \lvert p(z) \rvert \), and let \( z = re^{i\theta} \) where \( r = \lvert z \rvert \) and \( \theta \) is the argument of \( z \). Then:

\[ p(z) = a_0 + a_1re^{i\theta} + \ldots + a_{n-1}r^{n-1}e^{i(n-1)\theta} + r^ne^{in\theta} \]

\[ = a_0 + a_1re^{i\theta} + \ldots + a_{n-1}r^{n-1}e^{i(n-1)\theta} + r^ne^{i(n\theta)} \]

Now, let's consider the modulus of \( p(z) \):

\[ \lvert p(z) \rvert = \lvert a_0 + a_1re^{i\theta} + \ldots + a_{n-1}r^{n-1}e^{i(n-1)\theta} + r^ne^{i(n\theta)} \rvert \]

Using the triangle inequality, we have:

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