# Proving limits complex functions

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 $$1$$?

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