# 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 $f\left( z \right) = \frac{{i\bar z}}{2}$ in the open disk $\left| z \right| < 1$, then $\mathop {\lim }\limits_{z \to 1} f\left( z \right) = \frac{i}{2}
$
, 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 $z = \bar z$.

2. 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}
$
, 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

$z - 1 = x - 1 + iy$

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