An area enclosed by two arguments will be like a sector of a circle, but the circle would have, say, infinite radius. My dodgy way of describing it. Check the attachment for what I mean.
Oh. Maybe you don't know.
An argument, such as arg(z)=pi/4 is basically a 'line' which starts at origin, has gradient one (tan[pi/4]=1), and never ends. It's called a ray, as opposed to a line which has no visible starting point.
So pi/4 ≤ arg (z + i + 1) ≤ 3pi/4, this is two inequalities:
{arg(z+i+1)>=pi/4
{arg(z+i+1)<=3pi/4
Draw each seperately and then shade the region bounded by it (the least region in this case, because of the direction of the inequality signs). The ray is placed at -i-1, or (-1,-1). However, you exclude the case where you are taking the argument of zero: z+i+1=0, so exclude z=-1-i. (arg(0) undefined).
So you have two rays, both with same starting point, but one having a gradient of 1 and the other a gradient of -1.
Now the circle's the easy part.
Say hello, algebra bash (just to illustrate - it's easy to do it by inspection unless asked to use algebra)
|z + i + 1| ≤ 3
let z=x+iy, then:
|(x+1)+(y+1)i|<=3
sqrt[(x+1)^2 + (y+1)^2]<=3
(x+1)^2 + (y+1)^2 <= 9
Circle centre (-1,-1), radius 3, as well as everything inside it.
That help?
