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Use the formula

[tex]cos(x + iy) = cos(x)cos(iy) - sin(x)sin(iy)[/tex]

to find two imaginary numbers whose cosine is 3

My workings so far:

[tex]cos(x)cos(iy) - sin(x)sin(iy) = 3 [/tex]

Therefore:

[tex]cos(x)cosh(y) - i(sin(x)sinh(y)) = 3[/tex]

[tex]cos(x)cosh(y) = 3[/tex]

[tex]sin(x)sinh(y) = 0[/tex]

[tex]sin(x) = 0[/tex]

[tex]x = 0[/tex]

[tex]cos(x) = 1[/tex]

[tex]cosh(y) = 3[/tex]

[tex]y = ln(3 + 8^{\frac{1}{2}})[/tex]

However I can't seem to work out the other root. I know with polynomials it would would be its complex conjugate but I didn't want to assume so for other functions.