# Question #d266f

Oct 25, 2014

$| z - \left(2 + i\right) | < | z + 1 |$

by replacing $z$ by $x + i y$,

$R i g h t a r r o w | x + i y - \left(2 + i\right) | < | x + i y + 1 |$

by separating real parts and imaginary parts,

$R i g h t a r r o w | \left(x - 2\right) + i \left(y - 1\right) | < | \left(x + 1\right) + i y |$

by find the norms,

$R i g h t a r r o w \sqrt{{\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2}} < \sqrt{{\left(x + 1\right)}^{2} + {y}^{2}}$

by squaring,

$R i g h t a r r o w {\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2} < {\left(x + 1\right)}^{2} + {y}^{2}$

by multiplying out,

$R i g h t a r r o w {x}^{2} - 4 x + 4 + {y}^{2} - 2 y + 1 < {x}^{2} + 2 x + 1 + {y}^{2}$

by cleaning up a bit,

$R i g h t a r r o w - 2 y < 6 x - 4$

by dividing by -2,

$R i g h t a r r o w y > - 3 x + 2$

Hence, this is a region above the line $y = - 3 x + 2$.

I hope that this was helpful.