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Complex Equation

  1. Dec 15, 2005 #1
    Solve the following equations for the real variables x and y.
    (b) [tex]\left ( \frac{1 + i}{1 - i} \right )^2 + \frac{1}{x + iy} = 1 + i[/tex]
    I reduced this to
    [tex]2x + 2iy + ix - y - 1 = 0[/tex]
    but I cannot get any further. Have I reduced it correctly?
    Last edited: Dec 15, 2005
  2. jcsd
  3. Dec 15, 2005 #2


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    You have reduced it correctly.

    Your expression becomes,

    [tex](2x-y-1) + i(2y+x) = 0[/tex]

    The rhs is zero, so the lhs must be zero. And the only way that can happen is for both the real component and the imaginary component to be equal to zero.
  4. Dec 15, 2005 #3
    Thanks a lot, I got the answer now.
  5. Dec 15, 2005 #4


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    Your equation is equivalent to

    [tex]-1 + \frac{1}{x + iy} = 1 + i[/tex]


    [tex]\frac {1}{x + iy} = 2 + i[/tex]


    [tex]x + iy = \frac {1}{2 + i}[/tex]

    which, I think, is a little more straightforward.
  6. Dec 15, 2005 #5
    I've got another:

    Find the square root

    [tex](x + iy)^2 = -3 + 4i[/tex]

    I think I'm supposed to do some hand-waving here, but I would I truely work it out?
  7. Dec 15, 2005 #6


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    Have you learned to express complex numbers in polar form yet?
  8. Dec 16, 2005 #7
    Not yet, no.
  9. Dec 16, 2005 #8


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    Okay. In that case, just expand [itex](x+iy)^2[/itex] then equate real and imaginary parts on the left and right sides of the equation. This gives you two equations and two unknowns (x and y) to solve for. It's fairly straightforward. Let us know what you come up with.
  10. Dec 16, 2005 #9
    Doh, why didn't I think of that!? :mad:

    Anyway, I got the answer. Thanks!
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