0
$\begingroup$

On Pages 45-46 of Brown and Churchill's Complex Variables and Applications (Ninth Edition), we are given an example of a complex limit. The task is to show that $\lim_{z \to 1} f(z) = \frac{i}{2}$ for all $|z|<1$ where $f(z) = \frac{i\overline{z}}{2}$. He starts off by showing that $$\left|f(z) - \frac{i}{2}\right|=\left|\frac{i\overline{z}}{2} - \frac{i}{2}\right| = \frac{|z-1|}{2}.$$ From there, you can take $\delta = 2\epsilon.$ My question is, what happened to the $i$ in the inequality? Why can we factor it out and ignore it? Am I missing some properties of complex conjugates? Does it have to do with the fact that the modulus of these complex numbers is less than $1$?

$\endgroup$

2 Answers 2

1
$\begingroup$

Remember that absolute value is multiplicative: $|ab| = |a||b|.$ Thus, $|i\bar{z} - i| = |i||\bar{z} - 1| = |\bar{z} - 1|.$

The fact that $|z - c| = |\bar{z} - c|$ for $c \in \mathbb{R}$ follows by expanding. If $z = x+iy,$ then $|z-c| = \sqrt{(x-c)^2 + y^2}$ and $|\bar{z} - c| = \sqrt{(x-c)^2 + (-y)^2}.$

$\endgroup$
0
$\begingroup$

Sean answers it well. but remember that multiplying a complex number $z$ by $i$ results in $z$ rotated by 90 degrees. So $|iz| = |z|$. Since $iz$ is the same distance from the origin as $z$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .