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# Sin(z) watch

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1. If Sinz is bounded, analytic and entire in the complex plane why is it not constant?
2. (Original post by Goldenratio)
If Sinz is bounded, analytic and entire in the complex plane why is it not constant?
What makes you think it's bounded?
3. I'd say it's bounded by +-1 on the real line but i'm not sure about the complex plane.
4. (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)?
5. 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?
6. (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 as , 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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