Maths
posted by Tom .
use the formula
cos(x+iy)=cosxcosiysinxsiniy
to find two imaginary numbers whose cosine is 3
cos(x+iy)=cosxcosiysinxsiniy =
cos(x)cosh(y)  i sin(x)sinh(y) = 3
Equating imaginary parts gives:
sin(x)sinh(y) = 0
You know that y cannot be zero, otherwise the complex number woyuld be real, but for real arguments the cosine is always between 1 and 1. If y is not zero, sinh(y) is not zero, and therefore sin(x) must be zero. This means that x is an integer times pi.
x = n pi
Equating the real parts gives:
cos(x)cosh(y) = 3
Insert x = n pi, using
cos(n pi) = (1)^n
gives:
cosh(y) = 3*(1)^n
cosh for real arguments is always positive, so n must be even. And we see that:
y = arccosh(3)
So, the complex numbers are of the form:
2n pi + arccosh(3)i
Thanks
thanks
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