Solving $z^2 - 8(1-i)z + 63 - 16i = 0$ with reduced discriminant $z^2 - 8(1-i)z + 63 - 16i = 0$
I have come across a problem in the book Complex Numbers from A to Z and I do not understand the thought process of the author when solving:
The start to the solution is as follows:
$\Delta$ = discriminant
$\Delta ' = (4 - 4i)^2 - (63 - 16i) = -63 - 16i$
and $r = |\Delta '| = \sqrt{63^{2} + 16^{2}} = 65$, where $\Delta ' = (\frac{b}{2})^{2} - ac$
It is this last part I do not understand. I do not believe this definition for the discriminant has been previously mentioned in the book, and I struggle to see where it has come from.
Thanks

Note that solving quadratic equations involving complex numbers follows the same procedure as solving those involving real numbers.
Consider a quadratic equation with real coefficients: $$F(x)=ax^2+bx+c=0$$ How would you solve it? Well, the first step would involve solving out for the discriminant, right?
Let us now consider a quadratic with complex coefficients: $$F(z)=az^2+bz+c=0$$ As in a similar method like before, we can still calculate the Discriminant of $F(z)$ which equals, $\Delta = b^2-4ac$ and then you solve the quadratic.
What the author intended to do here was to find the Discriminant of the equation $az^2+2bz+c=0$ with $a=1$, $b=-4(1-i)$ and $c=63-16i$.
