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Roots of a complex equation

  1. Jan 19, 2006 #1
    find all solutions of the given equation: (z+1)^4=1-i
    im not sure if i did this right, but heres what i did
    the first thing that i did was notice that
    1-i = 2^1/2 * (cos (pi/4) + i*sin(pi/4))
    then i found
    z= [2^(1/2*1/4) * (cos (pi/4) + i*sin (pi/4) )^1/4] -1
    then using de moivre's thrm
    z= {2^3/4 *(cos [pi/16 +2k*pi/16] + i*sin[pi/16 + 2k*pi/16])}-1 ; k=0,1,2,3
    does this look right?
  2. jcsd
  3. Jan 20, 2006 #2
    Well the angle isn't pi/4 for the first thing. The point corresponding to 1-i is (1, -1)
  4. Jan 20, 2006 #3
    whoops, my bad
    1-i = (2^1/2) * [cos (-pi/4) + i*sin(-pi/4)]

    If i use this for 1-i and use the method of the first post will i obtain the correct answer?

  5. Jan 21, 2009 #4
    I just worked it out using your method and it looks good to me. You end up with four roots and they seem to correspond numerically to the values that Mathematica produces.
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