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# complex numbers, polynomials watch

1. I've seen this a couple of times used in proofs, and it's apparently so obvious it doesn't need to be proved itself. It's not obvious to be though...

write dD for the boundary of the unit circle in C. take f to be any polynomial. then:

sup_dD |z* - f(z)| >= 1

where z* is the complex conjugate, and >= means 'greater than or equal to'.

any thoughts?
2. I genuinely don't know if this would work, but how about the reverse triangle ineqality?

sup_dD |z* - f(z)| >= |z* - f(z)| >= | ||z*|| - ||f(z)|| | = | ||z|| - ||f(z)|| | = | 1 - ||f(z)|| |>= ... ?
3. I'm not seeing anything terribly obvious, but I think this works:

On dD, z* = 1/z. So |z*-f(z)| = |1/z - f(z)| = |z||1/z - f(z)| = |1 - z f(z)|.

Let g = 1-zf. g is a poly, so analytic. And |g(0)| = 1. Now apply the maximum modulus principle to g.

(There's probably a slightly more direct approach using Schwarz' Lemma).

by the ML-inequality

Edit: ninja'd by DFranklin
5. (Original post by SimonM)
I don't know how JohnnySPal's proof will work ...
Nor do I

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