Short answer: you are right.
Long answer:
First of all, note that for any integer n,
e^{i \theta} = e^{i (\theta + 2 \pi)}.
However, it is convention to normalize angles by subtracting multiples of 2 pi. Unfortunately two different conventions are in use: some people prefer to take all angles between -pi and pi, some prefer to use angles between 0 and 2 pi. If you are using the first [second] convention and you get a negative angle [angle > pi] you can always go to the other convention by adding [subtracting] 2pi.
Now, solving z4 = -i you first write z = r e^{i \theta}, i = e^{-i \pi}. The modulus equation gives r = 1, and then you get
e^{4 i \theta} = e^{-i \pi / 2}
(I am using the convention of angles between -pi and pi here, otherwise you would get 3pi/2 on the RHS).
The solution is given by
4 \theta = - \pi / 2 + 2 \pi n
so -- dividing by 4 --
\theta = - \frac{\pi}{8} + n \frac{pi}{2}
Now all you have to do is plug in values of n to get all the inequivalent angles between -pi and pi. You will find
4theta = -5*Pi/8, -Pi/2, 3*Pi/2, 7*Pi/2
If instead, you use the convention that angles should be between 0 and pi, you have to add 2 pi to the first two (i.e. add 8pi to 4 times the angle), and you get
4theta = 3*Pi/2, 7*Pi/2, 11*Pi/2, 15*Pi/2.
The book is apparently mixing the two conventions. Of course the answers are right, and if you want you could have written down
4theta = +75*Pi/2, -33*Pi/2, 3*Pi/2, -1593*Pi/2
if you wanted.