Finding and graphing the locus in the complex plane described by $1< |z-2| \leq 5$

217 Views Asked by At

To solve the inequality $1< |z-2| \leq 5$, the first thing I did was let $z=x+iy$ and managed to arrive at $1< (x-2)^2 + y^2 \leq 25$. Now this clearly represents the area between the circles with centre $(2,0)$ and radii 1 and 5. This seemed a bit odd to sketch so I tried to take away 1 from the inequality and arrived at $0< (x-2)^2 + y^2-1 \leq 25$ yet I wasn't sure if this was simply all the points inside the circle with centre $(3,-1) and radius 5 or if there was no representation of it all and my original idea was right.

1

There are 1 best solutions below

0
On BEST ANSWER

Your original idea was right. It is indeed the locus of points between two circles, of radii $r = 1, 5$, both centered in $(2, 0)$. To be precise, the locus includes only the circumference of the greater circle, since the first term of the associated inequality doesn't allow equality. You may graph the locus by plotting the lesser circle as a dotted circumference, as to show they're not included - they are only critical points of the locus.