Solution

Cauchy's formula gives

∣ f^{'} (z_{0}) ∣ \leq M_{R} / R

for an entire function bounded by

M

on every radius-

R

circle. If

f

is globally bounded, letting

R \to \infty

gives

f^{'} = 0

, proving Liouville's theorem.

Set

h (w) = f (1/ w)

. Its limit at zero makes the singularity removable, so

h (w) = a + c_{1} w + c_{2} w^{2} + \dots .

Consequently

f (z) = a + c_{1} / z + O (z^{- 2})

and

z^{2} f^{'} (z) \to - c_{1} =: b

The argument principle on a large circle gives number of zeros minus poles equal to the winding number of

g

, which is zero because

g \to 1

. Thus

m = n

G (z) = g (z) \frac{\prod _{i} ( z - p _{i} )}{\prod _{j} ( z - q _{j} )}

has removable singularities everywhere and tends to one at infinity. Liouville gives

G = 1

, hence

g (z) = \frac{\prod _{j} ( z - q _{j} )}{\prod _{i} ( z - p _{i} )} .

Solved by gpt-5.6-sol high.

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