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Real Part of a Function

  1. Oct 24, 2004 #1

    cj

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    I'm used to seeing a function like:

    [tex]\textbf{f}=x+iy \text [/tex]

    where

    [tex]i = \sqrt{-1}[/tex]

    and understanding the the real part is:

    [tex]\text{Re[f]} = x = Acos\theta[/tex]

    What, though, is the real part of a function like, for example,

    [tex]\textbf{f}=\sqrt{x+iy}[/tex]

    ??
     
  2. jcsd
  3. Oct 24, 2004 #2

    selfAdjoint

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    Expand in Taylor's series, paying attention to powers of i, and separate terms with i out to get two series for the real and imaginary parts. This only works where the TS converges absolutely. Where is that?
     
  4. Oct 24, 2004 #3

    robphy

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    [tex] Re(f)=\frac{f+f^*}{2}[/tex]
    where [tex]f^*[/tex] is the complex-conjugate of f.

    So, [tex]
    \begin{align*}
    Re(\sqrt{x+iy})&=\frac{(\sqrt{x+iy})+(\sqrt{x+iy})^*}{2}\\
    &=\frac{\sqrt{x+iy}+\sqrt{x-iy}}{2}
    \end{align*}
    [/tex]

    Additionally, you could write [tex]x+iy[/tex] in polar coordinates.
    Then, do the above.
     
  5. Oct 24, 2004 #4

    uart

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    Since it looks like you're assuming the x and y are real and using the notation x = A cos(theta), y= A sin(theta) then the answer is just,

    sqrt(A) cos(theta / 2)
     
    Last edited: Oct 24, 2004
  6. Oct 24, 2004 #5

    HallsofIvy

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    You need to be more careful in distinguishing between the "real and imaginary parts" of the variable z= x+ iy, and the "real and imaginary parts" of the function value. Most text books write f(z)= u+ iv so that u is the real part and v the imaginary part of f: is z= x+ iy then u(x,y) and v(x,y) are real valued functions of the real variables of x and y.
     
  7. Oct 24, 2004 #6

    cj

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    If I am (and I am) going to use polar forms, the
    general function:

    [tex]\textbf{f}=x+iy[/tex]

    has the magnitude:

    [tex]|\textbf{f}|=\sqrt{x^2+y^2}[/tex]

    with

    [tex]x = Acos\theta \text{ and } y = Asin\theta[/tex]

    [tex]\text{BUT, I have the form: } \textbf{f}=\sqrt{x+iy} \text{ , NOT }\textbf{f}=x+iy[/tex]

    So how do I make use of

    [tex]|\textbf{f}|=\sqrt{x^2+y^2}[/tex]

    since I don't have a function in the form of
    [tex]\textbf{f}=x+iy[/tex]

    ??
     
  8. Oct 25, 2004 #7

    HallsofIvy

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    Have you simply ignored what everyone has said?

    "I don't have a function in the form of f= x+ iy" is just saying you don't have the identity function: f(z)= z.

    You're not going to be able to do very much is f(z)= z is the only function you know how to work with!

    Suppose f(z)= z2. Then, writing z= x+iy, f(z)= (x+ iy)2=
    x2+ 2ixy+ y2(i2)= (x2- y2)+ (2xy)i.
    The real part of f is x2- y2 and the imaginary part of f is 2xy.

    f(z)= √(z) is a little harder just because it is harder to calculate the value. It would probably be best to write z in polar form:
    z= x+iy= r(cos(θ)+ i sin(θ)). The f(z)= √(z)= r1/2(cos(θ/2)+ i sin(θ/2)). The real part of f is r1/2cos(θ/2) and the real part is r1/2sin(θ/2).
     
  9. Oct 25, 2004 #8

    cj

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    Got it -- thanks!

    Although I think you meant "imaginary" rather than "real"
    in the following:


     
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