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Image of a complex function

  1. Nov 5, 2012 #1
    Let [itex]f(z)=z+\frac{1}{z}[/itex], the question is to find the image of this function on [itex]|z|>1[/itex].


    To do so, I tried to find the image of the unit circle which is the interval [-2,2] and so I could not determine our image.

    If also we tried to find the image of f we get
    [itex]f(re^{i\theta})=u+iv[/itex]
    where
    [itex]u(re^{i\theta})=(r+\frac{1}{r})\cos \theta[/itex]
    and
    [itex]v(re^{i\theta})=(r-\frac{1}{r})\sin \theta[/itex]
    with r>1.
     
  2. jcsd
  3. Nov 5, 2012 #2

    haruspex

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    No, [itex]f(re^{iθ})=re^{iθ}+e^{i(-θ)}/r[/itex]
    At r=1 this comes to the same, so the image of |z|=1 is indeed the [-2,2] interval of ℝ. But for |z| > 1 there will be an imaginary component.
    Looks to me like the image is still the whole of ℂ. As you increase r, the image becomes a series of concentric ellipses. A proof might be derivable by showing there are always real solutions to x+iy = z + 1/z.
     
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