How can one sketch a complex inequality with absolute values on both sides?
Solution 1:
Your calculation is correct. You now have to rewrite the inequality in the form:
$$(x-x_0)^2+(y-y_0)^2\ge r^2\;.$$
You will find:
$$x_0=-\frac{23}{12}$$
$$y_0=2$$
$$r=\frac{13}{12}\;.$$
Therefore, the region that satisfies the inequality is the part of the complex plane external ($\gt$) to the circle centred at $(x_0,y_0)$ of radius $r$ plus the circle itself ($=$).