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Homework Help: Real part of a complex number

  1. Sep 25, 2006 #1
    Hello there,
    I've been given the task of find the real part for the following expression

    [tex]\sqrt{x+iy}[/tex]

    And I'm a bit stuck. I figure that I'll just say that that equation is equal to some other imaginary number a+bi where 'a' is the real part and 'b' is the imaginary part, and try to solve for a. But after squaring both sides i get stuck immediately...

    [tex]x+iy = a^2 + 2abi - b^2[/tex]

    And i don't know where to go. Perhaps I'm going the wrong way with this one, any help would be appreciated.

    Edit: Ok, i've made some progress...

    I know then that

    [tex]x = a^2 - b^2[/tex]
    and
    [tex]iy = 2abi[/tex]

    So i try to get rid of the b term in the real one, but the only substitution I can make results in a y term being introduced into the real part, which is just adding another imaginary number in there...
     
    Last edited: Sep 25, 2006
  2. jcsd
  3. Sep 25, 2006 #2

    Hurkyl

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    Aren't you trying to express a and b in terms of x and y?

    No it's not -- y is not imaginary.
     
  4. Sep 25, 2006 #3
    ooh, after seeing iy i just labeled both as being imaginary, but i suppose that's not true...i'll keep working then!


    edit: i still seem to get stuck pretty quickly...

    making the substitution [tex]b=\frac{y}{2a}[/tex] and plugging that into [tex]x=a^2-b^2[/tex] gets me to:
    [tex]x=a^2- \frac{y^2}{4a^2}[/tex]
    And again, i feel stuck. I've tried putting them over a common denominator, but that doesn't seem to lead anyplace useful...
     
    Last edited: Sep 25, 2006
  5. Sep 25, 2006 #4

    HallsofIvy

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    Getting a common denominator is one way but the better way to handle equations with fractions is to multiply the entire equation by the "common" denominator- here just 4a2. If you do that you get the quartic equation [itex]4a^2= 4a^4- y^2[/itex] or [itex]4a^4- 4a^2- y^2= 0[/itex]. That's actually a quadratic equation in a2.
     
    Last edited by a moderator: Sep 26, 2006
  6. Sep 25, 2006 #5
    OK...tell me if this looks ok...it's really ugly and there's step or two which i'm not confident about:

    [tex]0=a^2-\frac{y^2}{4a^2}-x[/tex]
    [tex]0=\frac{4a^4 - 4xa^2 - y^2}{4a^2}[/tex]
    [tex]0=4a^4 - 4xa^2 - y^2[/tex]

    [tex]a^2 = \frac{4x \pm \sqrt{16x^2 - (4*4*-y^2)}}{2*4}[/tex]

    [tex]a^2 = \frac{4x \pm \sqrt{16x^2 + 16y^2}}{8}[/tex]

    [tex]a^2 = \frac{4x \pm \sqrt{16(x^2+y^2)}}{8}[/tex]

    [tex]a^2 = \frac{4x \pm 4\sqrt{x^2 + y^2}}{8}[/tex]

    [tex]a^2 = \frac{x \pm \sqrt{x^2 + y^2}}{2}[/tex]

    [tex]a=\sqrt \frac{x \pm \sqrt{x^2 + y^2}}{2}}[/tex]


    Can i leave it like that? (you may need to refresh the screen, i've fixed the tex)
     
    Last edited: Sep 25, 2006
  7. Sep 26, 2006 #6

    HallsofIvy

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    Yes, that looks good- except of course that you should have a [itex]\pm[/itex] on the outside and you don't really need the one inside the square root, only +. Taking the negative would make a imaginary and it must be real.

    You can check by looking at some simple examples. Suppose x is positive, y= 0. What does that formula give? Suppose x is negative, y=0. Suppose x= 0, y= 1. (The square root of i is [itex]\frac{\sqrt{2}}{2}(1+ i)[/itex] and [itex]-\frac{\sqrt{2}}{2}(1+ i)[/itex].)
     
    Last edited by a moderator: Sep 26, 2006
  8. Sep 26, 2006 #7
    alright, thanks for the pointers there, i can't say i've run accross many solutions with a root inside a root, seemed a bit odd to me...thanks again!
     
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