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Math Help - Complex numbers

  1. #1
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    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?

    Please, any help?
    Attached Thumbnails Attached Thumbnails Complex numbers-002.jpg  
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  2. #2
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    Using the properties from #4 above, notice that:

    \begin{gathered}<br />
  \overline {a\alpha ^2  + b\alpha  + c}  = \overline 0 \, \hfill \\<br />
  \overline {a\alpha ^2 }  + \overline {b\alpha }  + c = 0 \hfill \\<br />
  a\left( {\overline \alpha  } \right)^2  + b\overline \alpha   + c = 0 \hfill \\ <br />
\end{gathered}
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  3. #3
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    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}
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