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Square of complex numbers

  1. Dec 7, 2007 #1
    [tex]\sqrt{z}=\sqrt{r}(\cos \frac{\theta}{2} + \pi \ + \ \iota \sin \frac{\theta}{2} +\pi)\\[/tex] .....(i)
    Obtain from equation (i) this equation [tex] \sqrt{z} \ = \ +/- [\sqrt{\frac{1}{2} (\abs{z} + x)} \ + \ \mbox{ sign y} \ \iota \sqrt{\frac{1}{2} ( \abs{z}+x)}]\\ [/tex].
    Where sign y =1 it y greater than or equal to 0, sign y =-1 if y < 0 and all squares of positive numbers are taken with positive sign. And where on the rhs z=abs{z}.
    Equation (i) =[tex] - \sqrt{r}(\cos \frac{\theta}{2} \ + \ \iota \sin \frac{\theta}{2} )\\[/tex]
    => z = [tex] r(cos^2 \frac{\theta}{2} \ + \ 2 \iota \cos \frac{\theta}{2} sin\frac{ \theta}{2}\ + \ \iota^2 sin^2 \frac{\theta}{2}) \\[/tex]
    [tex] = r( \frac{1}{2}(1 \ + \ cos \theta ) \ + \ \iota sin \theta \ + \ \frac{1}{2}(1 \ - \ cos \theta))\\[/tex]
    [tex] = \frac{1}{2}r \ + \ \frac{1}{2} r cos\theta \ + \ \iota^2 \frac{r}{2} - \frac{1}{2} r cos \theta\\ [/tex]
    [tex] = \frac{1}{2} (\abs{z} \ + \ x) \ + \ \iota y \ - \ \frac{1}{2}r \ + \ \frac{1}{2} ( \abs{z} \ - \ x)\\ [/tex]
    THat is as far as I get in it. As you can see it is wrong. Note on the rhs z=abs{z}. Thanks for the help.
     
    Last edited: Dec 8, 2007
  2. jcsd
  3. Dec 7, 2007 #2

    Avodyne

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    Thre is a missing right parenthesis in your first expression. And the equation you are supposed to obtain has no equal sign and so is not an equation.
     
  4. Dec 8, 2007 #3
    The proper title is: Square root of complex numbers.
     
  5. Dec 8, 2007 #4

    Avodyne

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    You want to show that the two right-hand sides are equal. So why are you taking the square? And then you replace i^2 with 1 instead of -1 when going from the 1st to 2nd line of your "z =" equation. You should just get z = x + iy at the end of that. But it's not what you want to do in the first place.

    Your pi's in the first equation are also confusing. I assume that they are really part of the arguments of the sine and cosine, and that you just left out the parentheses. Also that they are optional; that's what would give you the plus-or-minus in the 2nd equation. It would help a lot if you would be precise in writing things down.
     
  6. Dec 9, 2007 #5
    First, I say sorry for my english, I will try write good.

    You can prove that with a algebraic method:

    We search a number z such a [tex]z^2=w[/tex], then, it's a system equations of form:

    [tex]x^2-y^2=u[/tex]
    [tex]2xy=v[/tex]

    If we raise to square both equations, and add the second equation to the fisrt equation:

    [tex]u^2+v^2=(x^2+y^2)^2[/tex]

    then we obtein a new system equation that we can solve easily:

    [tex]\sqrt{u^2+v^2}=x^2+y^2[/tex]
    [tex]x^2-y^2=u[/tex]

    I leave the rest for you, it is the key, raise to square and ad the equations.

    greeting from Grufey
     
  7. Dec 10, 2007 #6
    square root of complex numbers

    I squared it to get rid of a minus sign in front of [tex] \sqrt{r} \\[/tex]. The proper equation is: [tex] \sqrt{r}( \cos (\frac{\theta}{2} + \pi ) \ + \iota \sin( \frac{\theta}{2}+\pi)) \\ [/tex]. Thanks for reminding me about the need to be precise in maths.
    Welcome to the PF Grufey.
     
  8. Dec 12, 2007 #7
    I cannot follow Grufey's method he seems to be using two functions u and v, yet the section this question is in says nothing about functions. Later sections talk about complex functions.
     
  9. Dec 12, 2007 #8

    HallsofIvy

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    No, u and v are real numbers, the real and imaginary parts of [itex]z^2= (x+ iy)^2[/itex]
     
  10. Dec 12, 2007 #9
    Sorry, I left define [tex]z=x+iy[/tex] and [tex]w=u+iv[/tex]

    I recommended begin with [tex]z^2=w[/tex], I think that the problem is the notation, my z is my square root, I have a number w and I search a z such a [tex]z=\sqrt{w}[/tex], and you have z an you search w such a [tex]w=\sqrt{z}[\tex]. Sorry, I make a mistake with the notation.

    I hope that I can explain something.

    If you have some problem with complex analysis I recommend the book of Marsden Hoffman of complex analysis or ask in PF XD

    If you like I can do all. Do you like?
     
    Last edited: Dec 12, 2007
  11. Dec 13, 2007 #10
    Yes, Please do the question, as the book I use is not into giving examples, and I am short of examples.
     
  12. Dec 14, 2007 #11
    Ok, let's go:

    [tex]\sqrt{z}=\sqrt{r}(\cos \frac{\theta}{2} + \pi \ + \ \iota \sin \frac{\theta}{2} +\pi)\\[/tex]

    What is a square of a number z?. it is a number w such a the product with himself is equal to z. Then the last equation we can write as:

    [tex]\sqrt{z}=w[/tex]

    This is too general, because the second member of the equation is always a complex number that we define, w.
    Now we can raise to square:

    [tex]z=w^2[/tex]

    This is the equation that I said in other post. If [tex]z=z+iy[/tex] and [tex]w=u+iv[/tex] we have two equation, beacause, the fist is la equality of real parts, and the second equation is the equality of imaginary parts.

    [tex]x+iy=(u+iv)(u+iv)[/tex]

    then:

    [tex]u^2+v^2=x[/tex]
    [tex]2uv=y[/tex]

    we can raise to square both equations, and add the second equation to the first equation:

    [tex]x^2+y^2=(u^2+v^2)^2[/tex]

    The new system is:

    [tex]\sqrt{x^2+y^2}=u^2+v^2[/tex]
    [tex]u^2-v^2=x[/tex]


    [tex]u=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}[/tex]
    [tex]v=\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}[/tex]

    This two equation show 4 solutions, but the problem only have 2 solution, well, we can demostrate if we replace the solutions in the equation [tex]w^2=z[/tex] that the solution to the problem are:

    [tex]\sqrt{z}=w=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}+sign(v)\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}[/tex]
     
  13. Dec 14, 2007 #12
    Oh, the last equation, also you can see as:

    [tex]\sqrt{z}=w=\sqrt{\frac{x+\left|z\right|}{2}}+sign( v)\sqrt{\frac{-x+\left|z\right|}{2}}[/tex]
    or

    [tex]\sqrt{z}=w=\sqrt{\frac{Re(z)+\left|z\right|}{2}}+sign( v)\sqrt{\frac{-Re(z)+\left|z\right|}{2}}[/tex]

    This is trivial, but I say because you can see any time

    I only hope that you can bear my english, I have really problems to write and talk.

    greeting from Grufey
     
    Last edited: Dec 14, 2007
  14. Dec 15, 2007 #13
    Thanks for the help. Do you know who is the publisher of Marsden Hoffman's book.
     
  15. Dec 15, 2007 #14
  16. Dec 19, 2007 #15
    The problem you are trying to solve is called Demoivre Theorem, or you can solve it by applying Polar and Rectangular Cooordinates (a+bi)
     
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