Show that the minimum of a non-constant, non-zero analytic function on a closed bounded region occurs strictly on the boundary of the region.

Complex Analysis Problem



Problem.

(a) Let a function f be continuous on a closed bounded region R and analytic and not constant through-
out the interior of R. Assuming that f (z) = 0 anywhere in R, prove that |f (z)| has a minimum
value m in R which occurs on the boundary of R and never in the interior. Do this by applying the
corresponding result for maximum values to the function g(z) = 1/f (z).



(b) Use the function f (z) = z to show that in (a), the condition f (z) = 0 anywhere in R is necessary in
order to obtain the result of that exercise. That is, show that |f (z)| can reach its minimum value at
an interior point when the minimum value is zero.


Possible Solution.

For part (a), this theorem and corollary may be useful that state for f as described in (a), |f (z)| has no
maximum value in R but that the maximum always occurs on the boundary of R. Now if g = 1/f , we can
say that |g| also has no maximum so that for every z R, there is no z0 such that |g(z)| |g(z0 )| =
|1/f (z)| |1/f (z0 )| = |f (z)| |f (z0 )|. So this indicates that there is a minimum. Right?
Then what about (b)???