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COMPLEX SOLUTIONS help!

  1. Apr 27, 2008 #1
    First problem
    z^2+2z+1=0
    where the z in 2z is the conjugate (has a little line ontop)
    I just ignored the conjugate because I wasn't sure how to solve it, and I got -1 which is one of the solutions but there's also 1+2i and 1-2i which I understand because they're both conjugate of each other but I don't understand how they got it.
    Second problem
    z^3-3z^2+4z-12=0 given 2i is a solution... I don't even understand what they mean.
    Please help!
     
  2. jcsd
  3. Apr 27, 2008 #2
    As I see your problem is
    [tex]z^{2} + 2 z^{*} + 1 = 0[/tex]

    if you will search solution in following form [tex]z = a + i b[/tex], [tex]a, b[/tex] are both real numbers and insert it to your main equation then you will have system of two simple algebraic equations under [tex]a, b[/tex] and you'll find [tex]a = 1, b = \pm 2[/tex]. Solve your second problem in the same way and get the answer.
     
  4. Apr 27, 2008 #3
    I get that z=a+ib but how did you get values for a and b? I feel really stupid asking this but I dont see it. I tried solving it and then making b=0 and a=0 and I'm not getting 1 and 2 as values...
     
  5. Apr 27, 2008 #4
    What are your equations for a and b?
     
  6. Apr 27, 2008 #5

    HallsofIvy

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    Do the algebra. If z= a+ ib, then [itex]\overline{z}[/itex]= a- ib so [itex]z^2+ 2\overline{z}+ 1= (a+ ib)^2+ 2(a- ib)+ 1= 0[/itex]. Separate the real and imaginary parts and you have two equations for a and b.
     
  7. Apr 27, 2008 #6
    Yeah I got that far but i'm not sure how to seperate real and imaginary parts. Is it literally just placing all the real parts and making them equal to 0 and all the imaginary parts and make them equal to 0? Sorry our lecturer didn't go through this and so I'm just left lost. :shy:
    Thank you
     
  8. Apr 27, 2008 #7

    HallsofIvy

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    Yes, it literally is! If a+ bi= c+ di, then a= c and b= d. That's part of the definition of "complex number".
     
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