1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A complex number question

  1. Apr 3, 2007 #1
    i added a file in wich i tried to solve this question
    the final equation does"nt come out

    the question is (z-a)^3=8
    it is known that z1*z2*z3=-9

    i have dried to make an equation

    how do i solve this equation??

    plz help

    Attached Files:

    • comp.GIF
      File size:
      294.9 KB
  2. jcsd
  3. Apr 3, 2007 #2
    One way to approach the problem is to solve the first equation to find all possible values for (z-a). Since (z-a) is raised to the third power in this equation, we expect to find three possible values. This will give you three equations all of the form z-a=#.

    I'm not completely sure if I understand what you mean by the second equation, but I'm assuming that in your notation z1, z2, and z3 are the three possible values for z. So the three equations you got earlier are z1-a=#, z2-a=#, and z3-a=# (# stands for three different numbers). I'm also assuming that a is some complex constant. (Some particular number of the form x+y*i.)

    Using the three equations you got from solving the first equation and also using z1*z2*z3=-9 you can solve for z1, z2, z3, and a. This is because you now have four equations and four unknowns:
  4. Apr 3, 2007 #3
    the original (z-a)^3=8 has 3 roots
    z1 z2 z3
    i found them but transforming the equation to a trigonometric form
    and then i made DEMUAVER law and found these 3 roots
    then i multiplied them and i could not solve the equetion wich came out

    how do i solve this thing??
  5. Apr 4, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    I presume that when you used d'Moivre's formula you found that
    z- a= 2, or [itex]-1+ i\sqrt{3}[/itex], or [itex]-1-i\sqrt{3}[/itex].
    Then z= 2+ a, or [itex]a-1+i\sqrt{3}[/itex], or [itex]a-1- i\sqrt{3}[/itex].
    Multiplying those together and setting them equal to -9, you get a very simple equation for a!
  6. Apr 4, 2007 #5
    in the end of the file i showed the equation that came out
    its impossible to solve
  7. Apr 4, 2007 #6
    It's not impossible to solve. The complex numbers are algebraically closed.
  8. Apr 4, 2007 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    The equation you have on your file is exactly what I got: (a+2)(a2- 2a+ 4)= -9. Did you multiply it out? It is exactly a3+ 8= -9 or a3= -17. Surely you can solve that- just take the cube root.
  9. Apr 4, 2007 #8
    i didnt see that cube formula right under my nose
    it is so simple

    thank you very much
    Last edited: Apr 4, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: A complex number question