Thread: fp3 complex number

The number 8 question of the exercise says:
Use the formula
cos (x+iy)=cos x cos iy-sin x sin iy
to find two imaginary numbers whose cosine is 3
How do I use the formula here? I mean, why can't I just substitute (x+iy) in the place of z in the formula cos z =1/2 * (e^(iz)-e^-(iz))?

I noticed when I convert "cos x cos iy-sin x sin iy" in terms of e and simplify it leads to the same term as the substitution mentioned above, hence the question.

Originally Posted by ssadi

The number 8 question of the exercise says:
Use the formula
cos (x+iy)=cos x cos iy-sin x sin iy
to find two imaginary numbers whose cosine is 3
How do I use the formula here? I mean, why can't I just substitute (x+iy) in the place of z in the formula cos z =1/2 * (e^(iz)-e^-(iz))?

I noticed when I convert "cos x cos iy-sin x sin iy" in terms of e and simplify it leads to the same term as the substitution mentioned above, hence the question.

You want to solve set so .
Expand LHS and compare real and imaginary parts.