Proving Properties of Entire Functions | Cauchy's Theorem | Examples

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Homework Statement



Let f(z) be an entire function such that |f(z)| less that or equal to R whenever R>0 and |z|=R.

(a)Show that f''(0)=0=f'''(0)=f''''(0)=...

(b)Show that f(0)=0.

(c) Give two examples of such a function f.



Homework Equations





The Attempt at a Solution



I believe this has something to do with Cauchy but I not sure how to apply anything. Please I really need help with this one.

 
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I'm here for a bit, had apps to do.

All right, my advice is to first memorize the Cauchy inequalities for your final tomorrow (just memorize the actual inequality). For convenience, it says that if f is analytic on an open set U, and D(a,r) is a disk contained in U of radius r > 0 centered at a (advice: draw this out once), then

|f^{(n)}(a)| \leq \frac{n! M}{r^n}

where M is a bound for f on the boundary of the disk D(a,r) (i.e. |f(z)| \leq M for all z such that |z - a| = r).

Now in the problem, f is entire, so we don't need to worry about analyticity. For part a), what should play the role of a, M, r and n described in the theorem I just wrote?
 
a is 0 r>0 but i don't understand where M comes in exactly. how does this show that f''(0)=0=f'''(0) and so on

in addition do i show (b) and (c)?

i need this asap
 
M is a BOUND for f on the boundary circle. You really need to pay attention to the hypotheses in your problem. The first step in ANY problem is to translate the hypotheses.

The idea is that once we've established a bound on say |f''(a)|, we try to see if we can make it arbitrarily small according to the hypotheses. If we can, then |f''(a)| has to be 0, and that's part a). You can do the same for part b).

Part c) is not hard. It's easiest to think of polynomials which satisfy these properties (and surely you can come up with some, it's not a complex analysis question).
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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