Solution

Uniform convergence on compact subsets makes

f

continuous. For every triangle whose closure lies in

U

, uniform convergence on its boundary permits passage through the contour integral:

\int_{\partial Δ} f (z), d z = lim_{n \to \infty} \int_{\partial Δ} f_{n} (z), d z = 0.

f

is holomorphic. If a compact set

K ⋐ U

is surrounded by a finite union of contours at positive distance from

K

f_{n}^{'} (z) - f^{'} (z) = \frac{1}{2 πi} \int_{Γ} \frac{f _{n} ( ζ ) - f ( ζ )}{( ζ - z ) ^{2}}, d ζ .

The uniform bound on the surrounding compact set proves that

f_{n}^{'} \to f^{'}

uniformly on

K

f

had distinct zeros

c

and

d

, choose disjoint small closed discs around them whose boundary contains no zero of

f

. Uniform convergence and Rouché's theorem imply that, for large

n

f_{n}

has a zero in each disc, contradicting uniqueness of

c_{n}

. Thus

f

has at most one zero.

For an example on the unit disc, take

f_{n} (z) = z - (1 - 1/ n), c_{n} = 1 - 1/ n .

Then

f_{n} \to f (z) = z - 1

, which has no zero in the open disc. In general, Hurwitz's theorem shows that

f

is zero-free exactly when the unique zeros escape every compact subset of

U

for every compact K ⋐ U, c_{n} \in / K eventually .

Indeed, an interior accumulation point of

c_{n}

is a zero of

f

, while a zero of

f

forces the unique

c_{n}

into each of its sufficiently small neighbourhoods.

Solved by gpt-5.6-sol high.

 Ancestors (10)