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Given that the equation [tex]z^2+(p+iq)z+r+is=0,[/tex] where

  1. Jul 9, 2011 #1
    Given that the equation [tex]z^2+(p+iq)z+r+is=0,[/tex] where [tex]p,q,r,s[/tex] are real and non-zero root then
    which one is right
    1. [tex]pqr=r^2+p^2s[/tex]

    2. [tex]prs=q^2+r^2p[/tex]

    3. [tex]qrs=p^2+s^2q[/tex]

    4. [tex]pqs=s^2+q^2r[/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 9, 2011 #2

    SammyS

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    Re: Complexnumber

    What have you tried?

    Where are you stuck?

    To start, let z = x + y i , where x & y are real.

    I assume i 2 = -1
     
  4. Jul 10, 2011 #3
    Re: Complexnumber

    put [tex]z=x+iy[/tex][tex](x+iy)^2+(p+iq)(x+iy)+r+is=0[/tex] [tex]x^2-y^2+2ixy+px-qy+i(qx+py)+r+is=0[/tex][tex]\left(x^2-y^2+px-qy+r\right)+i\left(qx+py+2xy+s\right)=0+i.0[/tex][tex]\left(x^2-y^2+px-qy+r\right)=0[/tex][tex]\left(qx+py+2xy+s\right) = 0[/tex]

    Now I am struck here.
     
  5. Jul 10, 2011 #4

    SammyS

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    Re: Complexnumber

    Sorry for my lame suggestion !

    Use the quadratic formula to solve for z, then the part of the instructions which seem to be missing some words; in bold below.

    ... [itex]p,q,r,s[/itex] are real and non-zero root ...
     
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