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Math Help - Limit of a complex function

  1. #1
    Junior Member guildmage's Avatar
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    Limit of a complex function

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

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

    Observe that when z is in the disk \left| z \right| < 1, \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 z and each positive \varepsilon, \left| {f\left( z \right) - \frac{i}{2}} \right| < \varepsilon whenever 0 < \left| {z - 1} \right| < 2\varepsilon.

    We simply take \delta = 2\varepsilon.


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

    Isn't it that \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 f, but I don't know how this makes z = \bar z.
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  2. #2
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    Quote Originally Posted by guildmage View Post
    An example in proving the limit of a complex function says:

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

    Observe that when z is in the disk \left| z \right| < 1, \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 z and each positive \varepsilon, \left| {f\left( z \right) - \frac{i}{2}} \right| < \varepsilon whenever 0 < \left| {z - 1} \right| < 2\varepsilon.

    We simply take \delta = 2\varepsilon.


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

    Isn't it that \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 f, but I don't know how this makes z = \bar z.
    i\bar z = i(x - iy)

     = ix - i^2y

     = y + ix

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

     = y + i(x - 1).


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

    This is the same as |z - 1|, since

    z - 1 = x - 1 + iy

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