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Homework Help: [Complex variables] Solve: z^4 - 4z^2 +4 -2i = 0

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data
    "This is an example from my textbook:
    Solve the equation z4 - 4z2 + 4 - 2i = 0

    Rearranging, we get z4 - 4z2 + 4 = 2i
    or (z2 - 2)2 = 2i = (1+i)2
    This has solutions z2 - 2 = 1+i or -1-i.
    Equivalently z2=3+i or z2=1-i

    These may be solved to give the 4 solutions of the original equation.
    cv1.JPG "

    I don't understand the following step:
    (z2 - 2)2 = (1+i)2 => z2 - 2 = 1+i or -1-i

    Why is this true? I remember for real numbers we have √(x2) = |x| (note that it is |x|, not x). Is this true for complex numbers? If so, then
    (z2 - 2)2 = (1+i)2
    => z2 - 2 = +/- √[(1+i)2] = +/- |1+i| ???

    2. Relevant equations

    3. The attempt at a solution
    Shown above.

    I hope someone can explain this. Any help is appreciated!
    Last edited: Sep 23, 2010
  2. jcsd
  3. Sep 24, 2010 #2
    Make this substitution: z^2 = t
    and solve by the quadratic formula: t^2 - 4t + (4 - 2i) = 0
    get the answers, go back to the substitution and get in total 4 roots (some roots may be the same).
  4. Sep 24, 2010 #3


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    That is true for real numbers.

    However, you may have forgotten your equation solving: if you know that x2=a2, then x could be either a or it could be -a. In the real numbers, if you solved this equation by taking the (real) square root of both sides, you get |x|=|a|, not x=|a|.

    Anyways, in the complexes, x2 has two square roots, just as in the reals. Note that |x| usually isn't one of them. The principal square root (the analog of the real √ function) is the square root that either has positive imaginary part, or lies on the non-negative real axis. The other square root is its negative, as usual.

    Incidentally, even in the complexes, if you know that x2=a2, then it is also true that |x|=|a|. However, you can't do much with that, because there are lots of possible values for x, but only 2 of them satisfy the original equation
  5. Sep 25, 2010 #4
    I see.

    Then how can we actually justify or rigorously prove that
    (z2 - 2)2 = (1+i)2 => z2 - 2 = 1+i or -1-i ??
  6. Sep 25, 2010 #5


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    Because we know (1+i)2 has two square roots, and its obvious that those are them. Similarly, the two square roots of (z2-2)2 are obvious.

    Alternatively, bring them together and factor.
  7. Sep 25, 2010 #6
    But can we factorize/expand as usual when z and c are COMPLEX numbers??
    (z-c)(z+c) = z^2 - c^2

    If so, why?
  8. Sep 25, 2010 #7


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    Why can you do it for real numbers? The same reason applies here.
  9. Sep 26, 2010 #8
    For real numbers, it's true because we can expand it; I think it's called distributive law.

    Does this still hold for complex numbers z and c?
  10. Sep 26, 2010 #9


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    What you need to remember is not [itex]\sqrt{x^2}= |x|[/itex] (even for complex x, |x| is positive real number) but rather that the equation [itex]x^2= a^2[/itex] has the two roots x= a and x= -a. And, yes, that is true for complex numbers!

    (A little more about [itex]\sqrt{x^2}= |x|[/itex]. Working in the real numbers, we like functions to have one value for a given value of the argument- in fact, that is part of the definition of "function". That's why we define [math]\sqrt{a}[/math] to be the positive number whose square is a: [math]\sqrt{x^2}= |x|. But in the complex numbers we cannot do that- the complex numbers do not form an "ordered field" and, in particular, we cannot talk about "positive complex numbers". We have to accept "multi-valued functions" and, in fact, almost all functions, even those that are simple functions in the real numbers, become mult-valued functions over the complex numbers.)

  11. Sep 26, 2010 #10


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    The distributive law? Yes....

    (And the distributive law is easy to prove if you weren't sure about it)
  12. Sep 27, 2010 #11
    The method used in the example seems to apply only in very special cases.
    Is there a more general way to solve z4 - 4z2 + 4 - 2i = 0 ? Can we use the quadratic formula here? If so, can someone please outline the steps of solving this problem using the quadratic formula? (because I don't quite understand how to use it here and how we are supposed to get FOUR solutions this way)

  13. Sep 27, 2010 #12


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    The example solution is pretty much another form of the quadratic formula. They instead just completed the square. But if you want to try it out, just let u=z2 and solve for u, then substitute back
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