Help with proof in complex analysis

Question

I was looking at the proof of the result, the image of $\mathbb R_\infty$ under mobius transformation is a circle.

I don't follow how does this step

$(a\bar d-\bar bc)w+(b\bar c-\bar ad)\bar w+b\bar d-\bar bd=0$

Imply,this,

$2\,i\,\mathrm{Im}(a\bar d-\bar bc)w+b\bar d-\bar bd=0$

Answer 1

Hint:

$\overline{a+b}=\bar a+\bar b$ and $\overline{ab}=\bar a\cdot\bar b$, so

$$\overline{(a\bar d-\bar bc)w}=-(b\bar c-\bar ad)\bar w$$

And for any complex $z$,

$$z-\bar z=2\,i\,\mathrm{Im}\,z$$