Integration 497Note As we have seen in Section 5.14, in the special case of a polynomial
of degree n ;::: 1 the inequality (7.24-2) holds for all values of z of sufficientlylarge modulus; i.e., given M > 0 there exists R > 0 such that Jf(z)J > M
whenever Jzl > R.
Theorem 7.34 (Cauchy-Liouville Theorem for Nonanalytic Functions).
Suppose that1. f E '.D(C)
2. f 0 (z) =/:- 0 for all z E C and arbitrary (}
- Jf(z)J ~ M for all z E C
- limz-->oo f(z) = L
Then f is a constant function in C (E. R. Hedrick [17]).
Proof Clearly. L =/:-oo, and we may assume that L = 0 since otherwise it
suffices to consider the function fi(z) = f(z)-L. By defining f(oo) = O,
we have f(z) defined and continuous in the extended complex plane C*(a compact set). If f were not a constant in C* then Jf(z)J would attain
a nonzero maximum value at some finite point z 0 • By Theorem 6.28 this
is impossible.Remark If f~(z) = 0 for all z and all 0, then f is also a constant function
in C, as follows from Theorem 6.15. In fact, f = 0 since limz__, 00 f(z) = 0.However, if f 0 (z) = 0 at each z for some 0, then f need not be a constant.
Example Let z = reit and f(z) = e-zz = e-r
2
Then Jf(z)J ~ 1,
limz-->oof(z) = O, and
f~(z) = -e-zz(z + ze-2;0) = 0for z = 0 and(} arbitrary, or for z =/:- 0 and(} = t +^1 /i2k + l)rr (k any
integer).
7.25 Fundamental Theorem of Algebra
Theorem 7 .35 Every polynomial of degree n ;::: 1 has at least a zero.
Proof Let f(z) = aozn + aizn-l +···+an, ao =/:-0, n 2". 1. We know that
given M > 0, there exists R > 0 such that
Jf(z)J > M whenever JzJ > R (7.25-1)
Hence outside the circle Jzl = R f has no zeros. Suppose that on JzJ ~ R
f has no zeros either. Then g(z) = 1/ f(z) is analytic, thus continuous in
C, and g(z) =/:- 0 for all z E C. From (7.25-1) we have
1 1