Definition of a limit with complex variables

[MatheStudent]

Homework Statement

Use the definition of a limit to show that
$$lim (z^{2} +c) = z_{0}^{2} +c$$ as $${z->z_{0}}$$

Homework Equations

Definition of a limit:
|f(z)-L|< epsilon if
0<|z-z0|< delta

The Attempt at a Solution

|(z^{2}+c )-(z_{0}^{2}+c)| = | z^{2}-z_{0}^{2}|= |(z-z_{0})(z+z_{0}) | < epsilon
and I want to find a delta for |z-z0| but I don't know what to do with the z+z0 piece.
This is also my first time here, and I'm having problems getting this post looking nice.

 

[Fredrik]

This is also my first time here, and I'm having problems getting this post looking nice.

There's a bug that makes the wrong LaTeX images appear in previews. The only workaround is to refresh and resend after each preview, and sometimes you have to refresh one more time after saving the changes. You can edit your post during the first 11 hours and 40 minutes (700 minutes) after you posted it, but if you make major edits after someone replied, you should add a comment about it so that the guy who replied doesn't look like he's replying to something you never said.

Use itex tags around math expressions when there's text on the same line, and tex tags otherwise. (If the top gets cut off when you use itex, you might have to use tex even when there's text on the same line).

 

[Fredrik]

Homework Statement

Use the definition of a limit to show that
$$\lim (z^{2} +c) = z_{0}^{2} +c$$ as $$z\rightarrow z_{0}$$

Homework Equations

Definition of a limit:
|f(z)-L|< epsilon if
0<|z-z0|< delta

The Attempt at a Solution

$$|(z^{2}+c )-(z_{0}^{2}+c)| = | z^{2}-z_{0}^{2}|= |(z-z_{0})(z+z_{0}) | < \epsilon$$
and I want to find a delta for |z-z0| but I don't know what to do with the z+z0 piece.

I fixed some of your latex. (\lim and \rightarrow are useful codes). The first idea that occurs to me is to write $z+z_0=z-z_0+2z_0$. This way you get two terms, and you can try to choose $\delta$ to make each term $<\varepsilon/2$.

 

[MatheStudent]

Thanks for the help (in both latex and the problem), and I now have it solved, so thanks again.