Question

Let f be an entire function over the complex plane. Suppose
that
\[
|f(z)| \le 5 |z|, \,\, \text { for } \, |z|\ge 10
\]
What can you say about f?

Answer 1

ill say one, f is bounded above by 5|z| :p

Answer 2

You can actually show that f is a very specific function, like a polynomial of a certain degree.

Answer 3

is there a limit on the degree tho?

Answer 4

You have to show that it cannot exceed certain degree.

Answer 5

oh ye, sry misread thought u meant special function, would love to see it but no clue wat to start :P is there a answer somewhere?

Answer 6

Have you had complex analysis?

Answer 7

no, well i'll take a look later, ty

Answer 8

I am sorry for the newbie question, I didn't take any analysis course yet, but can you rewrite f(z) as f(a,b) = 5*(sqrt(a^2 + b^2)) if z = a + bi, and then the inequality becomes: sqrt(a^2 + b^2) >= 10. Don't know if this is the way to prove it, but I was wondering if you are allowed to change a complex-valued function with a real-valued function if you take the modulus of z.

Answer 9

You need to have a complex analysis course to deal with such a question.
Here a hint for people with the above background:
Notice that f(0) =0, then f(z) = z g(z) where g(z) is also entire,
Show that g(z) is bounded on the whole complex plane and conclude.

Answer 10

\[f(z).(f(z))^{*} \le 50\]
f(z)=x+yi->f(z).f(z)*=x²+y² <=50
so it image lies in a circle with radius sqrt(50)

Answer 11

This follows from Liouville's Theorem, second corollary on the wiki page:

en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis)

Answer 12

3rd corollary is related too.