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Complex solution of a cubic function.

  1. Dec 31, 2011 #1
    Hi,
    I'm studying my algebra and I was trying this exercise:

    Solve
    z³+iz²-7z-iz-6-6i=0

    I found the 2 real solutions (3 and -2) but i can't seem to find the complex one.

    I tried this:

    (z²-z)(i)=6(i) ==> z= -2 and 3
    z³-7z=6==> z= -1, -2 and 3

    I found that the complex solution is -1-i but i have no idea how to find it.

    (Im from Belgium, it's difficult to type maths in another language so sorry for mistakes)
     
  2. jcsd
  3. Dec 31, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well, the obvious thing to do is divide by z+ 2 and z- 3 so that you can factor that cubic into linear terms, then set each term equal to 0.
     
  4. Jan 2, 2012 #3
    Thanks, I found it
     
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