Help with complex numbers: az^2 + bz + c = 0; a, b, c real

[Guest]

Hi can i show that
a z^2 + b z + c = 0 where a,b, c are all real, must be of the form
z = x + iy, z = x-iy

 

[pka]

Re: Help with complex numbers

Hi can i show that
a z^2 + b z + c = 0 where a,b, c are all real, must be of the form
z = x + iy, z = x-iy

Please read this and tell what must be in that form!
Could it be the roots?
If so, what are we working in, the complex field?
Please expand your question to make it clear.

P.S. Because it is late for me:
\(\displaystyle \L
\begin{array}{l}
ax^2 + bx + c = 0,\quad \{ a,b,c\} \subseteq R \\
b^2 - 4ac \ge 0\quad \Rightarrow \quad x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
b^2 - 4ac < 0\quad \Rightarrow \quad x = \frac{{ - b}}{{2a}} \pm \frac{{\sqrt {\left| {b^2 - 4ac} \right|} }}{{2a}}i \\
\end{array}\)

 

[soroban]

Re: Help with complex numbers: az^2 + bz + c = 0; a, b, c re

Hello, americo74!

Can i show that the roots of: $\displaystyle \,az^2\,+\,bz\,+\,c\:=\:0$
where $\displaystyle \,a,b,c\,\in R$, must be of the form: $\displaystyle \,z \:= \:x\,\pm\,iy$

It's kind of obvious, isn't it?

The roots of any quadratic equation is: $\displaystyle \,z\;=\;\frac{-b\,\pm\,\sqrt{b^2\,-\,4ac}}{2a}$


And there are two cases to consider:

$\displaystyle [1]\;b^2\,-\,4ac\:\geq\:0$ . . . We have two real roots.

$\displaystyle \;\;\;$The roots are: $\displaystyle \:z\;=\;\left(\frac{-b\,\pm\,\sqrt{b^2\,-\,4ac}}{2a}\right)\,+\,0i$


$\displaystyle [2]\;b^2\,-\,4ac\: <\:0$ . . . We have two complex roots.

$\displaystyle \;\;\;$The roots are: $\displaystyle \:z\;=\;-\frac{b}{2a}\,\pm\,i\frac{\sqrt{4ac\,-\,b^2}}{2a}$