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Find the maximum value of $|f(z)|$ on the closed complex disk of radius 2,

where $f(z)$$=$$z^4\over{z^2+10}$.

Usually I approach these problems by calculating the modulus squared and simplifying, but here it seems it will slow things down a bit.

Johnny Apple
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1 Answers1

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Hint: first obtain a good (standard) bound on the largest $|f(z)|$ can be on $|z|=2$. Can you obtain this bound?

Evan
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  • Well, certainly the numerator is less than 16, but I'm am unsure about the denominator. I believe it has a lower bound of $|z|^2$ where $|z|=2$ so at 4. Thus an upper bound would be 4? – Johnny Apple Aug 31 '13 at 01:10
  • Use the reverse triangle inequality on the denominator. – Evan Aug 31 '13 at 01:11
  • So, -8/3 then? The top is 16, the bottom is -6. – Johnny Apple Aug 31 '13 at 01:20
  • Almost. A negative bound is pretty useless for a positive number :). But right, you got 6 on the denominator then. (you can use either order for that inequality). So which $z$ gets you 6 on the denominator? Whoops, I should also say that your negative bound is even wrong (dividing by negative reverses the direction of the inequality, so you would be saying $|f(z)| > $ something negative) – Evan Aug 31 '13 at 01:24
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    I'll assume you got that, the point is this is a simple method that avoids computation, because you know that you can do no better than the upper bound, so if you achieve it for some $|z|=2$ you are done. (Ok, minus the more pressing application as to why you can focus on $|z|=2$) – Evan Aug 31 '13 at 01:27
  • Well, so it suffices now to find a number which yield this reuslt? – Johnny Apple Aug 31 '13 at 01:29
  • @AnthonyVasaturo yes that's right – Evan Aug 31 '13 at 01:30