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By Jean Laurent

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**Example text**

Consider the sequence of disks of radius l, 2, 3, ... with centers at the coordinate origin. In each disk of radius k (for £ = 1 , 2 , . . ) , let zk denote a point at which the function attains its maximum in that disk: 28 ENTIRE FUNCTIONS \f&k)\ = M(k; /). Since Hm M{k\ f) = oo (by Liouville's theorem), the values k — OD of | / ^ ) | increases without bound. ±o at least for sufficiently large k. Furthermore, the absolute values \zk\ also approach oo. (Inside and on any circle of radius R, the absolute value of the function is bounded; therefore, the points z*at which \f(zk)\exceeds the maximum of l/fo) | inside and on that circle—which will be the case from some k on—must lie outside that circle).

General formula (53) shows that, in the f{z) is an wth-degree polynomial, the en also a polynomial of degree O^cn — k<^n. not vanish at z = a. The coefficient of the is an. 40 E N T I R E FUNCTIONS This result leads to a famous theorem of Bezout in the fol lowing form: If f(z) is a polynomial and z = a is a kth-order zero of f{z), then (z -— a)k divides f(z). Let us return to the general case of an entire function f(z) (which, as a special case, may be a polynomial). Let byfra denote an /th-order zero of the function f(z).

Then (cf. (*). where ® (z) is an entire function that does not vanish at z0. F r o m what we have just proven, t h e r e e x i s t s a disk with center at z0 in which y(z) does not vanish. Obviously, in that disk, f(z) has no z e r o s other than z{). This completes the proof of the l e m m a . F r o m this proposition, we derive the Theorem. An entire function f(z) not identically zero cannot have infinitely many zeros in any disk of finite radius. Proof (by contradiction): Suppose that the function f(z) has infinitely many zeros in the disk | z | ^ r .

### Approximation et optimisation by Jean Laurent

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