Question

use the formula
cos(x+iy)=cosxcosiy-sinxsiniy
to find two imaginary numbers whose cosine is 3

cos(x+iy)=cosxcosiy-sinxsiniy =

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