Results 1 to 5 of 5

Math Help - f(z) is derivable - then it's continuous - Prove it!

  1. #1
    Member
    Joined
    Sep 2010
    Posts
    95

    f(z) is derivable - then it's continuous - Prove it!

    Prove that if f is derivable in z_0 then f is continuous in z_0. \Omega \in \mathbb{C}, ~f:\Omega  \rightarrow \mathbb{C}, ~ z_0 \in \Omega

    I started with the definition.
    \lim_{z \to 0} \frac {f(z_0+\Delta z) - f(z_0)}{\Delta z}

    Using Cauchys integral formula.

    \frac {f(z_0+\Delta z) - f(z_0)}{\Delta z} =
    \frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)}{z-(z_0+\Delta z)} \, dz -\frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)}{z-z_0} \, dz =
    \frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)\Delta z}{(z-z_0-\Delta z)(z-z_0)} \,  dz =
    \frac{1}{2\pi i } \oint_C \frac{f(z)}{(z-z_0-\Delta z)(z-z_0)} \,  dz

    Now comparing the result with Cauchy's Integral formula, first order.
     \oint_C \frac{f(z)}{(z-z_0-\Delta z)(z-z_0)} \,  dz -  \oint_C \frac{f(z)}{(z-z_0)^2} \,  dz =  \oint_C \frac{f(z)\Delta z}{(z-z_0-\Delta z)(z-z_0)^2} \,  dz

    What happens with the right side if \Delta z \rightarrow 0?

    I'm sort of in the dark here... I thought maybe I proved something by showing how Cauchy's and the definition of derivation dance so nicely together.

    Help or any pointers would be appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    You're overcomplicating it! Note that lim_{z \to z_0} f(z)-f(z_0) = \lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0}\cdot (z-z_0)

    Also, I think that the proper term is 'differentiable', not 'derivable'.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2010
    Posts
    95
    Well as far as languages, I like Math and my native tongue. Bare with me... Tomorrow I'll dig into the stuff you suggested.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,976
    Thanks
    1121
    Another person who wants us to "bare" with him! This website is getting too raw for me!

    liquidFuzz, in order that a function be differentiable at z= a, you must have \lim_{\Delta z\to 0}\frac{f(a+ \Delta z)- f(a)}{\Delta z} exist. Since that denominator obviously goes to 0, in order that the limit exist, the numerator must also exist (a necessary though not sufficient condition). That is, we must have \lim_{\Delta z\to 0} f(a+ \Delta z}- f(a)= 0, whence \lim_{\Delta z\to 0} f(a+ \Delta z)= f(a). Letting z= a+ \Delta z that is the same as \lim_{z\to a} f(z)= f(a).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2010
    Posts
    95
    Mhmm, but by showing that the limit exists have I proved that a differentiable function is continuous?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. prove that every continuous function is separately continuous
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: November 23rd 2011, 03:57 AM
  2. Prove that f(x) is continuous.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2011, 01:35 PM
  3. Replies: 2
    Last Post: May 23rd 2011, 08:46 AM
  4. Replies: 6
    Last Post: November 17th 2009, 05:13 AM
  5. prove that f is derivable.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 23rd 2006, 08:57 AM

Search Tags


/mathhelpforum @mathhelpforum