+118609 Hi Solveit 
I let z=x+yi
and this is the graph I got.
I don;t intend to do all the working but I will say that
\(|z+4+3i|\\ =|(x+4)+(y+3)i|\\ =(x+4)^2+(y+3)^2\\ etc\)
see if you can do it from there,
+2498 So I came till here, but don't know how to continue:
\(|x+yi+4+3i|\leq|5\cdot(x+yi)+10-9i|\\ |(x+4)+i\cdot(y+3)|\leq|(5x+10)+i\cdot(5y-9)|\\ \sqrt{(x+4)^2+(y+3)^2}\leq\sqrt{(5x+10)^2+(5y-9)^2}\\ x^2+8x+16+y^2+6y+9\leq25x^2+100x+100+25y^2-90y+81\\ x^2+8x+y^2+6y+25\leq25x^2+100x+25y^2-90y+181\\ -24x^2-24y^2-92x+96y\leq156\)
