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Sin(z) watch

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    If Sinz is bounded, analytic and entire in the complex plane why is it not constant?:confused:
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    (Original post by Goldenratio)
    If Sinz is bounded, analytic and entire in the complex plane why is it not constant?:confused:
    What makes you think it's bounded?
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    I'd say it's bounded by +-1 on the real line but i'm not sure about the complex plane.
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    (Original post by Goldenratio)
    I'd say it's bounded by +-1 on the real line but i'm not sure about the complex plane.
    Well, what's your definition of sin(z)?
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    sin(z)=1/2i*(e^iz-e^-iz)

    hence this means that sinz is not bounded in the complex plane. Are we done? or should i say something else?
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    (Original post by Goldenratio)
    sin(z)=1/2i*(e^iz-e^-iz)

    hence this means that sinz is not bounded in the complex plane. Are we done? or should i say something else?
    Well rather than "hence", I think you should explictly say something like sin(ix)\to \infty as x\to\infty, so is unbounded. (It clearly wasn't obvious to you, so it's a bit disingenuous to just say "hence" without justification).

    As to whether we're done, it depends what you're trying to do I guess.
 
 
 
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Updated: August 6, 2007

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