#how do i solve for complex solutions for sin x = 3 and cos x = 3
winter cairnBOT
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lost panther
i'm not sure because i know the range only reaches 1 and -1 for sin x and cos x
twilit herald
use the complex definition of sine
$\frac{e^{iz}-e^{-iz}}{2i}=3$
in this case we set it equal to 3
dapper bayBOT
dontfwcrux
lost panther
$\frac{e^{iz}-e^{-iz}}{2i}=3$
where did you get this formula from?
twilit herald
where did you get this formula from?
uh
in the complex definition of e^iz
i want to say that there aren't any, because even complex numbers follow sin^2(x) + cos^2(x) = 1
twilit herald
i want to say that there aren't any, because even complex numbers follow sin^2(x...
there are
i have a general formula
somewhere
twilit herald
?
im not disproving your point, im just saying there are solutions
there can't be solutions if cos(x) = sin(x) = 3, as this will not follow the pythagorean identity
which is shown to hold true for complex numbers
oh thats what you mean
are you trying to solve them seperately or together?