Question

Question #6935c

Answer 1

Suppose we denote `z` by `z=x+iy` where `x,y in RR` then the locus of `z` will satisfy:

`| (x+iy)+(1+i) | = 2`

`=> | (x+1)+(y+1)i | = 2`

Then using the definition of the absolute value of a complex number, this requires that:

`sqrt( (x+1)^2+(y+1)^2) = 2`

`=> (x+1)^2+(y+1)^2 = 2^2`

So, `x,y` lie on a circle with centre `(-1,-1)` and radius `2`, which we can readily plot:
graph{(x+1)^2+(y+1)^2 = 2^2 [-5.98, 4.02, -3.48, 1.52]}

If we consider the the complex number `-1=2i` with associated coordinate `(-1,-2)` we can clearly see the coordinate lies within (but not on) the circle, we can prove this as follows:

With `x=-1` and `y=-2` we have:

`(x+1)^2+(y+1)^2 = (-1+1)^2+(-2+1)^2 = 1 < 2^2`

Showing that the coordinate does indeed lie within the circle, QED