How do I write arccos 4 as a + ib (complex number)?

This a question from complex number chapter of fp3

Find the following in the form a+ib, a,b member of real number set:

(a) arccos 4 (b) arcsin 2 (c) arcsin i

I tried to these sums by: x=arccos4, so cosx=4, so e^(ix)-e^-(ix)=8

The answers:

(a) My answer:+/- i arcosh 4

Book's answer: 2m*pi +/- i arcosh 4

(b) My answer: +/- arcosh 2(i am not sure how i get that)

Book's answer: (4m+1)*pi/2 +/- i arcosh 2

(c) My answer: i ln ((-2+/- sqrt 5)/2))

Book's answer: n*pi + i arsinh [(-1)^n]

So I gathered that I am missing something major here. Please anyone tell me the correct method to do these sums, thus enabling me to try it out myself. Just tell me the method, please:) This is my first sums of this kind:woo:

P.s. i=imaginary.