1. Complex numbers

Hi

For the circled question in the attachment, what exactly do I have to do to prove it? Do I let alpha = a + ib?

2. Using the properties from #4 above, notice that:

$\begin{gathered}
\overline {a\alpha ^2 + b\alpha + c} = \overline 0 \, \hfill \\
\overline {a\alpha ^2 } + \overline {b\alpha } + c = 0 \hfill \\
a\left( {\overline \alpha } \right)^2 + b\overline \alpha + c = 0 \hfill \\
\end{gathered}$

3. in every quadratic equation $a*(\alpha)^2 + b*(\alpha) + c = 0$

$\alpha_1,_2 = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$

or

$\alpha_1,_2 = \frac{ -b }{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

so if $b^2 - 4ac < 0$

than

$\sqrt{b^2 - 4ac} = \sqrt{(-1)*(b^2 - 4ac)} = i\sqrt{b^2 - 4ac}$ where you must see that $b^2 - 4ac$ on the left side is not the same on the right side

and now you have:

$\alpha_1,_2 = \frac{ -b }{2a} \pm i\frac{\sqrt{b^2 - 4ac}}{2a}$

if $f(\alpha) = *(\alpha)^2 + b*(\alpha) + c$

you need to find for what $\alpha$ where $f(\alpha) = 0$

but you know that that is for:

$\alpha_1,_2 = \frac{ -b }{2a} \pm i\frac{\sqrt{b^2 - 4ac}}{2a}$

and now you can see that:

$f( \bar{\alpha} ) = 0$

because:

$\bar{\alpha}_1,_2 = \frac{ -b }{2a} \mp i\frac{\sqrt{b^2 - 4ac}}{2a}$