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



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henpen
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I got \[y=0, x=2\pi n\]Correct?

hartnn
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seems correct.

henpen
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But are these the only solutions?

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

henpen
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\[\sin(x+iy)=\sin x \cosh y+i \cos x \sinh y\]

hartnn
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*because sin of...

henpen
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Cosine can cosh is a real function. It's imaginary for reals, however

hartnn
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cosine can never be imaginary
hyperbolic cos can be...i think this is your point...

henpen
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Yes. Thanks, I've got it now.