Results 1 to 3 of 3

Thread: Complex numbers

  1. #1
    Senior Member
    Joined
    Jul 2008
    Posts
    347

    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  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1
    Using the properties from #4 above, notice that:

    $\displaystyle \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} $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2009
    From
    Zagreb
    Posts
    65
    in every quadratic equation $\displaystyle a*(\alpha)^2 + b*(\alpha) + c = 0$

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

    or

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

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

    than

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

    and now you have:

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

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

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

    but you know that that is for:

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

    and now you can see that:

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

    because:

    $\displaystyle \bar{\alpha}_1,_2 = \frac{ -b }{2a} \mp i\frac{\sqrt{b^2 - 4ac}}{2a}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. raising complex numbers to complex exponents?
    Posted in the Advanced Math Topics Forum
    Replies: 10
    Last Post: Mar 25th 2011, 10:02 PM
  2. Replies: 1
    Last Post: Sep 27th 2010, 03:14 PM
  3. Imaginary numbers/complex numbers
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Aug 25th 2009, 11:22 AM
  4. Replies: 2
    Last Post: Feb 7th 2009, 06:12 PM
  5. Replies: 1
    Last Post: May 24th 2007, 03:49 AM

Search Tags


/mathhelpforum @mathhelpforum