P(z)=z^3 -3 + 2z -i = (z^3 – 3)+(2z- i)

a, |z-1|<1

0 < z < 2

Using Rouche Theorem, let z = 2, z^3-3= 5 > 2z - i, z^3 -3 is the dominant,

z^3-3 =0 and z has 3 roots.

So it means when z<2, P(z) has 3 roots.

let z = 0, z^3-3 = -3 < 2z - i, 2z - i is the dominant,

2z - i =0 and z has 1 root.

So it means when z<0, f(z) has 1 root.

So when 0 < z < 2, f(z) has 3-1 =2 roots.

b, |z-1|>1, |z|<2

z < 0, z > 2, -2 < z < 2 so -2 < z < 0

when -2 < z < 0, it has 1 root.

c, z > 2

when z < 2, it has 3 roots already, so when z > 2, it has 0 root.