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44), the term Pn(zk) cannot approach 0, whereas we have just shown that it and all the other terms must approach 0. This contradic tion proves the theorem. In particular, we may assert that ez, cos z, and sin z, are transcendental functions. 17. " In the first place, in the series f{z) = a0 + alz + aiz* + ... +anz*+ ... , we encounter terms of arbitrarily high powers of z with nonzero coefficients. In the second place, the maximum absolute value M (r; /) of such a function increases more rapidly than does the THE MAXIMUM ABSOLUTE VALUE 29 maximum absolute value of any polynomial no matter how high its degree.

This means that the complex roots of Eq. (70) are located symmetrically about the real axis; that is, they come in complex conjugate pairs. Therefore, we can speak of a sequence of roots for which ^ > 0 a n d another sequence (symmetric to it) for which # y<0. Let us begin by examining the roots for which y^>0. For these roots, we have, on the basis of the asymptotic equation proven above, Therefore, the second of Eqs. (70'") yields s i n ^ y ^ 1, and this means that y=~-\-2nK — en, where ert —0 as n—+ oo (with n assuming positive integral values).

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). It follows from formula (53) that the point b will also be a zero of the function y(z) since f(b) = (b-af9(b) = 09 and the inequality b — a ^ 0 implies that