# Thread: Continuity of Complex numbers

1. ## Continuity of Complex numbers

hello there,. i need your help with this continuity in complex plane problem.

Show that if a function f is continuous at a point z0 in some domain and f(z0) not equal to 0, then there exists a neighborhood of z0 throughout which f(z) not equal to 0.

thnx

2. ## Re: Continuity of Complex numbers

Originally Posted by aldrincabrera
hello there,. i need your help with this continuity in complex plane problem. Show that if a function f is continuous at a point z0 in some domain and f(z0) not equal to 0, then there exists a neighborhood of z0 throughout which f(z) not equal to 0.
Choose $\epsilon=|f(z_0)|$ .

3. ## Re: Continuity of Complex numbers

the book suggested me with this,.
write the continuity as |f(z0) - f(z)|/2 where epsilon = |f(z0)|/2. then note the contradiction if f(z)=0 at some point z in every neighborhood of z0.

now i am confused,.

4. ## Re: Continuity of Complex numbers

Originally Posted by aldrincabrera
the book suggested me with this,.
write the continuity as |f(z0) - f(z)|/2 where epsilon = |f(z0)|/2. then note the contradiction if f(z)=0 at some point z in every neighborhood of z0.
Suppose that $|f(w)-f(z_0)|<\frac{|f(z_0)|}{2}$.

Then it follows that $0<\frac{|f(z_0)|}{2}<|f(w)|$.

Can we have $0<|f(w)|=0~?$.

5. ## Re: Continuity of Complex numbers

no??what should i do next??my mind is messed up

6. ## Re: Continuity of Complex numbers

Originally Posted by aldrincabrera
no??what should i do next??my mind is messed up
You are done, at no $w\in \mathcal{B}_{(f(z_0)/2)}$ can $f(w)=0$.

7. ## Re: Continuity of Complex numbers

,.sir,.what will happen if f(w) = 0??will there have any contradiction??

8. ## Re: Continuity of Complex numbers

Originally Posted by aldrincabrera
,.sir,.what will happen if f(w) = 0??will there have any contradiction??
If $|f(w)|>0$ then $f(w)\ne 0$.

9. ## Re: Continuity of Complex numbers

Originally Posted by aldrincabrera
the book suggested me with this,. write the continuity as |f(z0) - f(z)|/2 where epsilon = |f(z0)|/2. then note the contradiction if f(z)=0 at some point z in every neighborhood of z0. now i am confused,.
Of course we can choose $\epsilon=|f(z_0)|/2$ but it is also valid if $\epsilon=|f(z_0)|$ . As $f$ is continuous at $z_0$ there exists $\delta >0$ such that $|f(z)-f(z_0)|<|f(z_0)|$ if $|z-z_0|<\delta$ .

If $f(z)=0$ then, $|f(z)-f(z_0)|=|f(z_0)|<|f(z_0)|$ (contradiction).