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Complex mapping

  1. Apr 28, 2006 #1
    Hi everyone - I'm sure there's somebody here who can help with such a trivial question.

    It's not a homework question before that's assumed - it's from a past exam paper, which I'm using for revision, sans answers.

    It asks to describe, in the complex plane, the image of:

    [tex] |z - 1| \leq 1 [/tex]

    under the transformation

    [tex] w = \frac{1}{z} [/tex]

    Now, I found the answer to be:

    [tex] Re(w) \leq \frac{1}{2} [/tex]

    Is this correct?

    Thanks for any help,

    Simon.
     
  2. jcsd
  3. Apr 28, 2006 #2

    HallsofIvy

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    Are you sure about the direction of that inequality? z= 1 itself satisfies |z-1|=0< 1 and z= 1 is mapped into w= 1 which has real part 1> 1/2.
     
  4. Apr 28, 2006 #3
    Oops - no, I'm not sure of that at all - what I actually meant (of course!) was

    [tex] Re(w) \geq \frac{1}{2} [/tex]
     
  5. Apr 28, 2006 #4

    HallsofIvy

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    Yep. If z= x+ iy, then [itex]w= \frac{1}{z}= \frac{1}{x+iy}\frac{x-iy}{x-iy}= \frac{x-iy}{x^2+y^2}[/itex] which has real part [itex]\frac{x}{x^2+ y^2}[/itex].

    If z satisfies [itex]|z-1|\le 1[/itex] then [itex]|z-1|= \sqrt{(x-1)^2+ y^2}\le 1[/itex] so [itex](x-1)^2+ y^2= x^2+ y^2- 2x+ 1\le 1[/itex].
    That is, [itex]x^2+ y^2\le 2x[/itex] and therefore [itex]\frac{x}{x^2+ y^2}\ge \frac{1}{2}[/itex].
     
  6. Apr 28, 2006 #5
    Excellent - thanks very much for your help.

    Simon.
     
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