Question

Complex analysis problem. Help needed

Answer 1

Prove that the function f(z)=1/z^2 is not uniformly continuous in the region |z|<=1 but it is uniformly continuous in the region 1/2 <=|z|<=1.

Answer 2

impossible to divide by zero...
so when z = 0
f(z)=1/z^2
f(0)=1/0^2
f(z)=1/0
f(z)=infinity / not rational

Answer 3

does that make sense?

Answer 4

@Jack1 we cannot put z=0 and do the problem. We need to show it by
\[\epsilon- \delta\] process

Answer 5

@Kainui

Answer 6

@Michele_Laino

Answer 7

I think that the subsequent inequalities, can be useful:
\[ \left| {\frac{1}{{{z^2}}} - \frac{1}{{z_0^2}}} \right| = \frac{{\left| {z_0^2 - {z^2}} \right|}}{{\left| {{z^2}z_0^2} \right|}} \leqslant \frac{{\left| {z_0^2 - {z^2}} \right|}}{{2\left| {z_0^2} \right|}} \leqslant \frac{{2\left| {z - {z_0}} \right|\left| {z + {z_0}} \right|}}{{2\left| {z_0^2} \right|}} \leqslant \frac{\delta }{{2\left| {{z_0}} \right|}} \leqslant \varepsilon \]
Of course the absolute values symbol \(|.|\) stands for the distance in \(\mathbb{C}\)

Answer 8

oops.. absolute value* symbol...