solving complex number inequality

Sep 2013
567
27
Portland
Inequalities have always gotten me for whatever reason

$|z+i| \le 3$

well, $z=x+iy$

so it becomes $|x+iy+i| \le 3 $ which is $|x+i(y+1)| \le 3$

$\sqrt{x^2+(y+1)^2} \le 3$

$-9 \le x^2 + y^2 + 2y +1 \le 9$

$-9 - 2y -1 \le x^2 +y^2 \le 9 - 2y -1$ I think Im just complicating it
 

Plato

MHF Helper
Aug 2006
22,474
8,643
Inequalities have always gotten me for whatever reason
$|z+i| \le 3$
That is set of points that compose a closed circular disk centered at $-\bf{i}$ with radius 3.
 
Sep 2013
567
27
Portland
yeah I know that it is, maybe I should have posted about a more difficult inequality in my problems. I am supposed to sketch the points. so the points would be $(-3,-i)$ and $(3,-i)$ correct?

EDIT: nevermind I sketch the entire disk. duh
 
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