Limit of a complex function

**guildmage** · December 14th 2009, 05:31 PM

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 $z = \bar z$ .

**Prove It** · December 14th 2009, 05:38 PM

Originally Posted by guildmage

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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