# Limits in the complex vectorspace

## Recommended Posts

Hi,

I am trying to prove that a limit exists at a point using the epsilon delta definition in the complex plane, but I can't seem to reach a conclusion.

Here's what I have been trying to get at:

[LATEX]\lim_{z\to z_o} z^2+c = {z_o}^2 +c[/LATEX]

[LATEX] |z^2+c-{z_o}^2-c|<\epsilon \ whenever\ 0<|z-z_o|<\delta[/LATEX]

[LATEX]LH=|z^2-{z_o}^2|=|z-z_o||z+z_o|[/LATEX]

[LATEX]=|z-z_o||\overline{z+z_o}|[/LATEX]

[LATEX]=|z-z_o||\bar{z}+\bar{z_o}|[/LATEX]

[LATEX]=|z\bar{z} +z\bar{z_o} -{z_o}\bar{z} -z_o\bar{z_o}|[/LATEX]

[LATEX]=| |z|^2 -|z_o|^2 +2Im(zz_o) |[/LATEX]

[LATEX]\leq||z|^2 -|z_o|^2 +2|z||z_o|| \ (because\ Im(z)\leq|z|)[/LATEX]

But I can't get any further. I did this much thinking I could factor it to the square of delta, but that didn't work out because of the positive 2zzo term.If anyone can help me out here, it would be great. Thanks.

##### Share on other sites

Hi,

I am trying to prove that a limit exists at a point using the epsilon delta definition in the complex plane, but I can't seem to reach a conclusion.

Here's what I have been trying to get at:

[LATEX]\lim_{z\to z_o} z^2+c = {z_o}^2 +c[/LATEX]

[LATEX] |z^2+c-{z_o}^2-c|<\epsilon \ whenever\ 0<|z-z_o|<\delta[/LATEX]

[LATEX]LH=|z^2-{z_o}^2|=|z-z_o||z+z_o|[/LATEX]

[LATEX]=|z-z_o||\overline{z+z_o}|[/LATEX]

[LATEX]=|z-z_o||\bar{z}+\bar{z_o}|[/LATEX]

[LATEX]=|z\bar{z} +z\bar{z_o} -{z_o}\bar{z} -z_o\bar{z_o}|[/LATEX]

[LATEX]=| |z|^2 -|z_o|^2 +2Im(zz_o) |[/LATEX]

[LATEX]\leq||z|^2 -|z_o|^2 +2|z||z_o|| \ (because\ Im(z)\leq|z|)[/LATEX]

But I can't get any further. I did this much thinking I could factor it to the square of delta, but that didn't work out because of the positive 2zzo term.If anyone can help me out here, it would be great. Thanks.

You are overly complicating the problem. |z+z0| < (2+δ)|z0| for |z-z0| < δ. You should be able to work out the δ, ε relationship - remember z0 is fixed.

Edited by mathematic
##### Share on other sites

• 4 weeks later...

Hey, how exactly did you reach |z+z0| < (2+δ)|z0| for |z-z0| < δ.

I can't seem to understand..

##### Share on other sites

|z + z0| = |z - z0 + 2z0| ≤ |z - z0| + 2|z0| < δ + 2|z0|

##### Share on other sites

I forgot I put this up...

Thanks mathematic !

## Create an account

Register a new account