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Exact Cartesian Form

  1. Nov 4, 2007 #1
    Hello,
    My question comes in two parts, I don't know if the first part is relevent to the second so i'll put it in anyway.

    a. Express 1 + root(3)i in polar form

    I can solve this to get:

    2cis(pi/3)

    My problem is with part b.

    b. Solve the quadratic equation z^2 + 2z - root(3)i = 0, expressing your answers in exact cartesian form

    I used the quadratic formula (I don't like completing the square) to get:

    z = (-2 + root(4 + 4root(3)i))/2 and z = (-2 -root(4 + 4root(3)i))/2

    However, i'm lost with the 'exact cartesian form' part.

    Any help would be appreciated, thanks.
     
    Last edited: Nov 4, 2007
  2. jcsd
  3. Nov 4, 2007 #2

    Dick

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    I think cartesian form just means express the answer in the form a+bi. You now have to express the square roots of the complex quantities in that form. It might actually have been easier to complete the square.
     
  4. Sep 3, 2009 #3
    Ok I think You should get the real part and the imaginary part:
    I think this is:
    -2/2 and -2/2 are the reals.
    Sqrt(4+4Sqrt3i)/2 and -Sqrt(4+4Sqrt3i)/2
    Put these together and you should get a real part and an imaginary part.
    I hope that helps. Don't take this as the real answer I might be wrong.
    Check with your teacher or your tutor or whoever.
    I'm pretty sure thats right, I'll keep thinking........hmmmm....
     
  5. Sep 3, 2009 #4
    wait I see an error.
    (-2 + root(4 + 4root(3)i))/2
    The 4+4root.... is wrong it should be 4-4root...
    Remeber the equation is -b+/- root(b^2-4ac)/2a
    Try again it might work.
     
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