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B No distinction between positive and negative for imag nums?

  1. Aug 1, 2016 #1
    I am solving for the square root of a complex number, namely: ##\sqrt{5 -12i}## Let's assume for brevity's sake that I already have the solution ##3- 2i##. This is fine and well, but the book that I am reading goes on to say that "##- (3 -2i)## is an equally valid solution, since there is no distinction between 'positive' and 'negative' for nonreal numbers." How is this so? Why do both of these solution work for the root of a complex number, while if I were calculating ##\sqrt{5 - 12 \sqrt{2}}## there would only be one solution, namely the positive one?
     
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  3. Aug 1, 2016 #2
    This is clearly untrue. Just consider the example ##w = a + 0i## and ##z = -(a + 0i)##. Clearly there is a distinction.

    Now, for example, in the case of the reals, there is a difference between finding the square root of a number and solving for the roots of a polynomial. If I wrote "what is ##\sqrt{4}##?" then the answer would be ##2##, but if I asked for the roots of ##x^2 - 4 = 0 \Rightarrow x^2 = 4##, then there are clearly two solutions, ##-2## and ##2##.

    But this is just common practice. When people write the square root symbol with no context, they mean the positive solution; however, clearly the negative solution squared will yield the term in the square root as well.
     
  4. Aug 1, 2016 #3

    jbriggs444

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    But neither i nor -i are positive. There is no algebraic test that can distinguish between ##i=\sqrt{-1}## and ##j=-\sqrt{-1}##
     
  5. Aug 1, 2016 #4

    pwsnafu

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    The quote says "nonreal numbers". You need to give examples from ##\mathbb{C}\setminus\mathbb{R}##.
     
  6. Aug 2, 2016 #5

    Svein

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    When you are in the complex domain there are new rules and new ways to look at things. Example: When you say ##5 -12i## I immediately convert it to ##r\cdot e^{i\phi} ## with r being 13 and φ being -1.17601. From there you can take the square root easily - (square root of r and φ/2). But in doing so, you forget that ##e^{i\phi}## and ##e^{i(\phi+2\pi)} ## represent the same number! So, taking the square root you must include both the solution with φ/2 and the solution with φ/2+π. And, since ##e^{i\pi}=-1 ##, you see why the other solution is included...
     
  7. Aug 2, 2016 #6

    ehild

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    The n-th root of an imaginary number Z is not unique. It means all the numbers z1, z2 z3 ..... the n-th power of which is Z: (zk)n=Z.

    You can find the second root of 5-12i as a+ib so as (a+ib)2=5-12i. This means real equations for both the real part and imaginary part:
    a2-b2=5, 2ab=-12. Eliminating b and solving the quadratic equation for a2, only one solution is positive, a2=9, so the solutions are a = 3 and b = -2; and a = -3, b = 2.

    As for the complex n-th root, the n-the roots of the unit 1 are cos((2π/n)k)+isin((2π/n)k) for k=0, 1, n-1, or, in Euler's notation e((2π/n)ki). So the second roots of 1 are cos(0)+i sin(0) = 1 and cos(π) +i sin(π) = -1.
    The n-th roots of a general complex number ##Z=|Z|e^{iΦ}## are ##\sqrt[n] {|Z|}e^{i(Φ/n + (2π/n) k }## for k=0, 1, 2...n-1 where |Z|1/n is the real n-th root of |Z|.
    Written in trigonometric form, ##Z^{1/n}=\sqrt[n] {|Z|}(\cos(Φ/n + (2π/n) k)+i\sin(Φ/n + (2π/n) k)##.
    This way the second roots of Z=5-12i are = √13 (cos(Φ/2+πk)+i sin( Φ/2+πk) ), where Φ=arctan(-12/5) cos(Φ)=5/13, sin(Φ)=-12/13, and k = 0, 1.
     
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