How is this step done? $\left|\frac{i\overline{z}}{2} -\frac i2\right|=\frac{|z-1|}{2}$ Absolutely everything makes sense other than what is in red. How is this step completed? 

Let us show that if $f(z)=\dfrac{i\overline{z}}{2}$ in the open disk $|z|\lt 1$, then$$\lim \limits_{z\to1}f(z)=\frac i2.$$
  That the point $1$ being on the boundary of the domain of $f$. Observe that when $z$ is in the disk $|z|\lt 1,$
$$\left|f(z)-\frac i2\right|=\color{red}{\left|\frac{i\overline{z}}{2} -\frac i2\right|=\frac{|z-1|}{2}}.$$
Hence for any such $z$ and each positive number $\epsilon$,
$$\left|f(z) -\frac i2\right|\lt \epsilon$$ whenever $$0\lt |z-1| \lt 2\epsilon.$$

My attempt to resolve:
$$i\overline{z} - i=i(a-bi)-i=ai+b-i=(a-1)i+b\ne (a+bi-1)$$

I believe the intermediate steps are $$\left|\frac{i\bar{z}}{2} - \frac{i}{2}\right| = |i|\left|\frac{\bar{z}}{2} - \frac{1}{2}\right| = 1\frac{|\bar{z} - 1|}{2} = \frac{1}{2}\left|\overline{z - 1}\right| = \frac{1}{2}|z - 1|$$
where the last equality follows from the fact that $|\bar{a}| = |a|$ for all $a \in \mathbb{C}$.
