Tiberius Posted March 28, 2012 Posted March 28, 2012 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.
mathematic Posted March 28, 2012 Posted March 28, 2012 (edited) 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 March 28, 2012 by mathematic
Overlord_Prime Posted April 22, 2012 Posted April 22, 2012 Hey, how exactly did you reach |z+z0| < (2+δ)|z0| for |z-z0| < δ. I can't seem to understand..
mathematic Posted April 22, 2012 Posted April 22, 2012 |z + z0| = |z - z0 + 2z0| ≤ |z - z0| + 2|z0| < δ + 2|z0|
Tiberius Posted April 23, 2012 Author Posted April 23, 2012 I forgot I put this up... Thanks mathematic !
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