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Thread: Simplify complex equation

  1. #1
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    Simplify complex equation

    The question:
    $\displaystyle |\sqrt{w^2 - 4} + w| = 2$ such that w is a complex number

    I tried using the identity $\displaystyle |z|^2 = z\overline{z}$ to get a solution, but it appears to create a mess. :/

    Any assistance would be great.
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  2. #2
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    $\displaystyle \displaystyle |\sqrt{w^2 - 4} + w| = 2$

    $\displaystyle \displaystyle |\sqrt{w^2 - 4} + w|^2 = 2^2$

    $\displaystyle \displaystyle (\sqrt{w^2 - 4} + w)^2 = 4$.

    Now try to solve for $\displaystyle \displaystyle w$...
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  3. #3
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    Quote Originally Posted by Prove It View Post
    $\displaystyle \displaystyle |\sqrt{w^2 - 4} + w| = 2$

    $\displaystyle \displaystyle |\sqrt{w^2 - 4} + w|^2 = 2^2$

    $\displaystyle \displaystyle (\sqrt{w^2 - 4} + w)^2 = 4$.

    Now try to solve for $\displaystyle \displaystyle w$...
    I'm not sure this will work since $\displaystyle |a|^2 = a \overline{a}$, not a^2, when a is complex.

    @OP: I'd be inclined to substitute w = x + iy and take it from there, messy as it's likely to be .... (Where has the question come from?)
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    Quote Originally Posted by mr fantastic View Post
    I'm not sure this will work since $\displaystyle |a|^2 = a \overline{a}$, not a^2, when a is complex.

    @OP: I'd be inclined to substitute w = x + iy and take it from there, messy as it's likely to be .... (Where has the question come from?)
    It's part of a complex analysis question:

    Show that the function f defined by $\displaystyle f(z) = z^2 + z^{-2}$ maps the unit circle onto the interval [-2, 2]
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    Quote Originally Posted by Glitch View Post
    It's part of a complex analysis question:

    Show that the function f defined by $\displaystyle f(z) = z^2 + z^{-2}$ maps the unit circle onto the interval [-2, 2]
    If $\displaystyle z$ is on the unit circle then $\displaystyle |z|^2 = z \overline{z} = 1$, so $\displaystyle z^{-1} = \overline{z}$.

    Then $\displaystyle f(z) = z^2 + \overline{z}^2 = (z + \overline{z})^2 - 2 z \overline{z} = (z + \overline{z})^2 -2$

    So...
    Last edited by awkward; Mar 19th 2011 at 12:46 PM. Reason: typo
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    Quote Originally Posted by awkward View Post
    If $\displaystyle z$ is on the unit circle then $\displaystyle |z|^2 = z \overline{z} = 1$, so $\displaystyle z^{-1} = \overline{z}$.
    Sorry, I'm not sure how you came to the conclusion that $\displaystyle z^{-1} = \overline{z}$.
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  7. #7
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    Because on the unit circle, $\displaystyle \displaystyle |z|^2 = 1 \implies z\overline{z} = 1 \implies \overline{z} = \frac{1}{z}$.
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  8. #8
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    Oh, right. >_<

    I need stronger coffee.
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  9. #9
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    Quote Originally Posted by awkward View Post
    $\displaystyle (z + \overline{z})^2 -2$
    OK, so I've worked out what you've done so far. This is what I've done:

    Let z = x + iy

    $\displaystyle (x + iy + x - iy)^2 - 2$
    $\displaystyle (2x)^2 - 2$
    $\displaystyle 4x^2 - 2$

    So unless I'm mistaken, this is a parabola. I'm fairly sure this isn't the correct solution. :/
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    Quote Originally Posted by Glitch View Post
    OK, so I've worked out what you've done so far. This is what I've done:

    Let z = x + iy

    $\displaystyle (x + iy + x - iy)^2 - 2$
    $\displaystyle (2x)^2 - 2$
    $\displaystyle 4x^2 - 2$

    So unless I'm mistaken, this is a parabola. I'm fairly sure this isn't the correct solution. :/
    Since (x,y) is on the unit circle, $\displaystyle -1 \leq x \leq 1$.

    So then what is the range of $\displaystyle 4x^2 - 2$?
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  11. #11
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    Ahh, thanks.
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