# Thread: Limit of a complex function

1. ## 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 $\displaystyle z = \bar z$.

2. Originally Posted by guildmage
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

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

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