Solving $z^2 - 8(1-i)z + 63 - 16i = 0$ with reduced discriminant

Question

$z^2 - 8(1-i)z + 63 - 16i = 0$

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

Answer 1

Generally if you have and equation of the form

$$az^2+2bz+c=0$$

Then $\Delta= (2b)^2-4ac = 4(b^2-ac)= 4 \Delta'$

Definition: $\color{blue}{\Delta'=b^2-ac}$ is the so called the reduced discriminant of the equation $az^2+2bz+c=0$.

Therefore the solution are

$$z=\frac{-2b\pm\sqrt{\Delta}}{2a} =\color{red}{\frac{-b\pm\sqrt{\Delta'}}{a}} $$

if you have

$$az^2+bz+c=0$$ then, $$\color{brown}{\Delta' =\left(\frac{b}{2}\right)^2-ac}$$

Answer 2

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 $c=63-16i$.

Answer 3

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.