From this, the two real solutions are:
(a, b) = (√3, -√3) and (-√3, 3),
giving the two square roots of √(-6i) as √3 - i√3 and -√3 + i√3; it doesn't matter which one of these we use to substitute back into the quadratic formula. For the sake of choice I am choosing √3 - i√3 as the square root of √(-6i).
So, e^(iz) = [-1 + 3i ± √(-6i)]/4 now becomes:
e^(iz) = [-1 + 3i ± (√3 - i√3)]/4,
giving e^(iz) = [(-1 + √3) + (3 - √3)i]/4 and e^(iz) = [(-1 - √3) + (3 + √3)i]/4 as solutions.
At this point, taking the natural logarithm of both sides and dividing by i gives two possible solutions to the equation; all others can be obtained by adding multiples of 2πi to these (since e^z is a multi-valued function with period 2πi).
(2) While it is true that -1 ≤ sin(x) ≤ 1 for all real x, the same cannot be said for complex x; in fact, sin(x) can take any complex number we want if we consider complex values of x.
As to how to solve sin(z) = 4, it is solved in the same manner as the first problem, but probably isn't as messy.
I hope this helps!