G
Guest
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
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
Elite Member
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- Jan 29, 2005
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Re: Help with complex numbers
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:
\(\displaystyle \L
\begin{array}{l}
ax^2 + bx + c = 0,\quad \{ a,b,c\} \subseteq R \\
b^2 - 4ac \ge 0\quad \Rightarrow \quad x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
b^2 - 4ac < 0\quad \Rightarrow \quad x = \frac{{ - b}}{{2a}} \pm \frac{{\sqrt {\left| {b^2 - 4ac} \right|} }}{{2a}}i \\
\end{array}\)
Please read this and tell what must be in that form!americo74 said:
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:
\(\displaystyle \L
\begin{array}{l}
ax^2 + bx + c = 0,\quad \{ a,b,c\} \subseteq R \\
b^2 - 4ac \ge 0\quad \Rightarrow \quad x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
b^2 - 4ac < 0\quad \Rightarrow \quad x = \frac{{ - b}}{{2a}} \pm \frac{{\sqrt {\left| {b^2 - 4ac} \right|} }}{{2a}}i \\
\end{array}\)
Re: Help with complex numbers: az^2 + bz + c = 0; a, b, c re
Hello, americo74!
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: \(\displaystyle \:z\;=\;-\frac{b}{2a}\,\pm\,i\frac{\sqrt{4ac\,-\,b^2}}{2a}\)
Hello, americo74!
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: \(\displaystyle \:z\;=\;-\frac{b}{2a}\,\pm\,i\frac{\sqrt{4ac\,-\,b^2}}{2a}\)