<s>[INST] 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
 [/INST] Since $∆' = -63-16i$
$$ z = \frac{ -b±\sqrt{∆'}}{a} $$
$$ z= 8 - 8i ± \sqrt{-63-16i} $$

Using Quadratic Formula, instead of reduced determinant

$$ z= \frac{8(1-i)±\sqrt{\left[8(1-i)\right]^2-4\cdot1\cdot(63+16i}}{2\cdot1} $$
$$ = \frac{8-8i±\sqrt{\left[64(2i)\right]-\cdot(4×63+4×16i}}{2} $$
$$ = \frac{8-8i±\sqrt{\left[(128i)\right]-(192+64i}}{2} $$
$$ = \frac{8-8i±\sqrt{128i-192-64i}}{2} $$
$$ = \frac{8-8i±\sqrt{192+64i}}{2} $$
$$ = \frac{8-8i±\sqrt{16(12+4i)}}{2} $$
$$ = \frac{8-8i±4\sqrt{4(3+i)}}{2} $$
$$ = \frac{8-8i±8\sqrt{(3+i)}}{2} $$
$$ = \frac{8(1-i)±8\sqrt{(3+i)}}{2} $$
$$ = 4(1-i)±4\sqrt{(3+i)} $$
$$ = 4[(1-i)+\sqrt{(3+i)}],  4[(1-i)-\sqrt{(3+i)}]$$
Are two possible solutions of your equation.
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