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henpen

  • 2 years ago

Solve \[\sin(x+iy)=0\]

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  1. henpen
    • 2 years ago
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    I got \[y=0, x=2\pi n\]Correct?

  2. hartnn
    • 2 years ago
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    seems correct.

  3. henpen
    • 2 years ago
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    But are these the only solutions?

  4. hartnn
    • 2 years ago
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    they are already infinite....and u know y must be 0 cos sin of imaginary number can't be real

  5. henpen
    • 2 years ago
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    \[\sin(x+iy)=\sin x \cosh y+i \cos x \sinh y\]

  6. hartnn
    • 2 years ago
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    *because sin of...

  7. henpen
    • 2 years ago
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    Cosine can- cosh is a real function. It's imaginary for reals, however

  8. hartnn
    • 2 years ago
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    cosine can never be imaginary hyperbolic cos can be...i think this is your point...

  9. henpen
    • 2 years ago
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    Yes. Thanks, I've got it now.

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