Results 1 to 2 of 2

Thread: Limit of a complex function

  1. #1
    Junior Member guildmage's Avatar
    Joined
    Aug 2009
    From
    Philippines
    Posts
    35

    Limit of a complex function

    An example in proving the limit of a complex function says:

    Let us show that if $\displaystyle f\left( z \right) = \frac{{i\bar z}}{2}$ in the open disk $\displaystyle \left| z \right| < 1$, then $\displaystyle \mathop {\lim }\limits_{z \to 1} f\left( z \right) = \frac{i}{2}
    $, the point $\displaystyle 1$ being on the boundary of the domain of definition of $\displaystyle f$.

    Observe that when $\displaystyle z$ is in the disk $\displaystyle \left| z \right| < 1$, $\displaystyle \left| {f\left( z \right) - \frac{i}{2}} \right| = \left| {\frac{{i\bar z}}{2} - \frac{i}{2}} \right| = \frac{{\left| {z - 1} \right|}}{2}$.

    Hence, for any such $\displaystyle z$ and each positive $\displaystyle \varepsilon$, $\displaystyle \left| {f\left( z \right) - \frac{i}{2}} \right| < \varepsilon$ whenever $\displaystyle 0 < \left| {z - 1} \right| < 2\varepsilon$.

    We simply take $\displaystyle \delta = 2\varepsilon$.


    I understand most parts of the proof. However, I don't know how they got to $\displaystyle \left| {\frac{{i\bar z}}{2} - \frac{i}{2}} \right| = \frac{{\left| {z - 1} \right|}}{2}$.

    Isn't it that $\displaystyle \left| {\frac{{i\bar z}}{2} - \frac{i}{2}} \right| = \frac{{\left| i \right|\left| {\bar z - 1} \right|}}{2} = \frac{{\left| {\bar z - 1} \right|}}{2}$?

    I'm guessing it's because of the domain of definition of $\displaystyle f$, but I don't know how this makes $\displaystyle z = \bar z$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by guildmage View Post
    An example in proving the limit of a complex function says:

    Let us show that if $\displaystyle f\left( z \right) = \frac{{i\bar z}}{2}$ in the open disk $\displaystyle \left| z \right| < 1$, then $\displaystyle \mathop {\lim }\limits_{z \to 1} f\left( z \right) = \frac{i}{2}
    $, the point $\displaystyle 1$ being on the boundary of the domain of definition of $\displaystyle f$.

    Observe that when $\displaystyle z$ is in the disk $\displaystyle \left| z \right| < 1$, $\displaystyle \left| {f\left( z \right) - \frac{i}{2}} \right| = \left| {\frac{{i\bar z}}{2} - \frac{i}{2}} \right| = \frac{{\left| {z - 1} \right|}}{2}$.

    Hence, for any such $\displaystyle z$ and each positive $\displaystyle \varepsilon$, $\displaystyle \left| {f\left( z \right) - \frac{i}{2}} \right| < \varepsilon$ whenever $\displaystyle 0 < \left| {z - 1} \right| < 2\varepsilon$.

    We simply take $\displaystyle \delta = 2\varepsilon$.


    I understand most parts of the proof. However, I don't know how they got to $\displaystyle \left| {\frac{{i\bar z}}{2} - \frac{i}{2}} \right| = \frac{{\left| {z - 1} \right|}}{2}$.

    Isn't it that $\displaystyle \left| {\frac{{i\bar z}}{2} - \frac{i}{2}} \right| = \frac{{\left| i \right|\left| {\bar z - 1} \right|}}{2} = \frac{{\left| {\bar z - 1} \right|}}{2}$?

    I'm guessing it's because of the domain of definition of $\displaystyle f$, but I don't know how this makes $\displaystyle z = \bar z$.
    $\displaystyle i\bar z = i(x - iy)$

    $\displaystyle = ix - i^2y$

    $\displaystyle = y + ix$

    So $\displaystyle i\bar z - i = y + ix - i$

    $\displaystyle = y + i(x - 1)$.


    $\displaystyle |i\bar z - i| = \sqrt{y^2 + (x - 1)^2}$.

    This is the same as $\displaystyle |z - 1|$, since

    $\displaystyle z - 1 = x - 1 + iy$

    $\displaystyle |z - 1| = \sqrt{(x - 1)^2 + y^2}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limit of complex function
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Nov 14th 2010, 09:58 PM
  2. limit from sides in complex function..
    Posted in the Calculus Forum
    Replies: 14
    Last Post: Jul 12th 2010, 01:16 PM
  3. Limit of complex function
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: Oct 7th 2009, 07:55 PM
  4. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Sep 3rd 2009, 05:05 PM
  5. Replies: 1
    Last Post: Mar 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum