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henpenBest ResponseYou've already chosen the best response.1
I got \[y=0, x=2\pi n\]Correct?
 one year ago

henpenBest ResponseYou've already chosen the best response.1
But are these the only solutions?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
they are already infinite....and u know y must be 0 cos sin of imaginary number can't be real
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[\sin(x+iy)=\sin x \cosh y+i \cos x \sinh y\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Cosine can cosh is a real function. It's imaginary for reals, however
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
cosine can never be imaginary hyperbolic cos can be...i think this is your point...
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Yes. Thanks, I've got it now.
 one year ago
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