Complex Numbers Inequality

Homework Statement

Determine the region in the complex plane described by |z-2i| < |z+ i|

Homework Equations

z= x+ iy
|z|= (x2 + y2)1/2

The Attempt at a Solution

|z-2i| < |z+ i|

|z-2i|/|z+ i| < 1

|z-2i| = [(x-2i)2 + y2]1/2
|z+ i| = [(x+i)2 + y2]1/2

[(x-2i)2 + y2]1/2
--------------- < 1
[(x+i)2 + y2]1/2

[(x-2i)2 + y2]1/2*[(x+i)2 - y2]1/2
---------------------------------- < 1
[(x+i)2 + y2]1/2*[(x+i)2 - y2]1/2

Am I on the right track of solving this so far? If so how do I proceed to the next step? If not what part did I do wrong? Any feedback is appreciated!

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Ray Vickson
Homework Helper
Dearly Missed

Homework Statement

Determine the region in the complex plane described by |z-2i| < |z+ i|

Homework Equations

z= x+ iy
|z|= (x2 + y2)1/2

The Attempt at a Solution

|z-2i| < |z+ i|

|z-2i|/|z+ i| < 1

|z-2i| = [(x-2i)2 + y2]1/2
|z+ i| = [(x+i)2 + y2]1/2

[(x-2i)2 + y2]1/2
--------------- < 1
[(x+i)2 + y2]1/2

[(x-2i)2 + y2]1/2*[(x+i)2 - y2]1/2
---------------------------------- < 1
[(x+i)2 + y2]1/2*[(x+i)2 - y2]1/2

Am I on the right track of solving this so far? If so how do I proceed to the next step? If not what part did I do wrong? Any feedback is appreciated!
|z-i2|^2 is NOT (x-2i)^2 + y^2. Think about why not.

RGV

dynamicsolo
Homework Helper
Trying to compute the inequality in Cartesian coordinates is an efficient way to make yourself crazy.

It may be more helpful to use a geometrical interpretation of the equation first in the Argand diagram. Keep in mind that | z - z0 | is the "length" of a vector from the point representing z0 to the point representing z . The equation | z - 2i | = | z + i | then describes the curve in the Argand diagram of points equidistant from 2i and -i . What is that curve like? The inequality then represents the set of points closer to 2i than to -i . Where is that region?