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Complex Number problems.

  1. Oct 24, 2003 #1


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    Heyas. I'm not too good at complex numbers so excuse me if these questions are a bit on the laughable side.

    Find all real values for r for which ri is a solution of the equation.
    z^4 - 2z^3 + 11z^2 - 18z + 18 = 0
    hence, Determine all the solutions of the equations...

    I'm not really sure how to tackle questions like this. I subbed in 'ri' for 'z' and worked out r, but it had multiple values, some of which are not solutiuons as listed in the back of the book. Perhaps i did it wrong all together.

    A quick guide through the question would help heaps :)


    A similar question
    Find the real number 'k' such that z=ki is a root of the equation:
    z^3 + (2+i)z^2 + (2+2i)z + 4 = 0.

    is this basically the same as the earlier question?

    Thanks in advance :)
  2. jcsd
  3. Oct 24, 2003 #2


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    Science Advisor

    Okay, assuming you are given that there is an imaginary solution,
    replace z with ri. Then, as I am sure you calculated, z^2= -r^2,
    z^3= -r^3 i and z^4= r^4. Substituting those into the equation,
    r^4+ 2r^3i- 11r^2- 18ri+ 18= 0.

    Of course, in order for this to be 0, both real and imaginary parts must be 0. The real part is r^4- 11r^2+ 18= 0 and the imaginary part is 2r^3- 18r= 0. The crucial part is that the same r must satisfy BOTH equations. It's easy to factor the second equation:
    2r^3- 18r= 2r(r^2- 9)= 0. The solutions are r=0, r= 3 and r= -3. It's obvious that r=0 does not satisfy r^4- 11r^2+ 18= 0. What about r= 3 and r= -3?

    Now, for z^3 + (2+i)z^2 + (2+2i)z + 4 = 0.

    Again, if z= ri, then z^2= -r^2 and z^3= -r^3 i. Substituting that into the equation, -r^3 i+ (2+i)(-r^2)+ (2+ 2i)(ri)+ 4=
    -r^3 i- 2r^2- r^2 i+ 2ri- 2r+ 4= 0.

    Separate real and imaginary parts: the real part is
    -2r^2- 2r+ 4= 0. The imaginary part is -r^3- r^2+ 2r= 0.
    The first of those factors as -2(r+2)(r-1)= 0 and so has solutions r= 1, r= -2. The second factors as -r(r^2+ r- 2)= -r(r+2)(r-1)= 0 and so has solutions r= 0, r= 1, r= -2. Remembering that r must satisfy BOTH equations, what can r be?
  4. Oct 24, 2003 #3


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    Thanks for that man :)
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