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Homework Help: 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


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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


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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:


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


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


    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


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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:


    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:


    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.




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


    The new system is:



    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:

  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]

    [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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