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Homework Help: Complex Numbers

  1. Oct 2, 2006 #1
    2 Questions:

    (1.)Carry out the indicated calculation:

    [tex](\frac{-6+2i}{1-8i})^2[/tex]

    =[tex]\frac{36-24i+4i^2}{1-16i+64i^2}[/tex]

    since [tex]i^2=-1[/tex]

    =[tex]\frac{32-24i}{-63-16i}[\frac{-63+16i}{-63+16i}][/tex]

    I carry out the math and get an answer of:

    [tex]\frac{-2400+1512i}{4225}[/tex]

    I must be doing something wrong with the conjugate, as far as i can tell it looks right, but for some reason my signs should be switched to give me the correct answer of:

    [tex]\frac{-1632+2024i}{4225}[/tex]

    (2.)determine the set of all z satisfying the given equation or inequality:

    [tex]|z-2i|<=|z+1+i| & |z|> 4[/tex]

    I solved this one down to:
    [tex]x^2+y^2-4y+4<=x^2+y^2+2x+2y+2[/tex] & [tex]x^2+y^2>4[/tex]

    what do I do to simplify it?

    Thanks.
     
  2. jcsd
  3. Oct 2, 2006 #2

    radou

    User Avatar
    Homework Helper

    Well, you just have to find the intersection of [tex]y \geq -\frac{1}{3}x+\frac{1}{3}[/tex] (implied by the first inequality) and [tex]x^2+y^2 > 4[/tex], i.e. [tex]S=\left\{(x,y) \in \textbf{R}^2: y \geq -\frac{1}{3}x+\frac{1}{3} \wedge x^2+y^2 > 4 \right\}[/tex]. Draw a sketch of S.
     
    Last edited: Oct 2, 2006
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