Complex Circle Comparison Comparing Complex Circles: Solving |z+2| < |z+2i|

[MaxManus]

Homework Statement

Hey, hope that someone can be nice and help me describe this set:

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

The Attempt at a Solution

z = x + iy
|z+2| < |z+2i|
sqrt((x+2)^2 + y^2) < sqrt(x^2 + (y+2)^2)
possible to do more?

The left side is the equation for a circle with center x = -2, y = 0 and the right side is the equation for a circle with center x = 0 and y = -2.

So the set is the circle described on the left side with radius less than the circle on the right side.
Is this correct?
And if I want to sketch the set?

 

[LCKurtz]

Homework Statement

Hey, hope that someone can be nice and help me describe this set:

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

The Attempt at a Solution

z = x + iy
|z+2| < |z+2i|
sqrt((x+2)^2 + y^2) < sqrt(x^2 + (y+2)^2)
possible to do more?

Yes, it is possible to do more. Simplify it. Start by using the fact that if a and b are nonegative numbers and a < b, then a²<b².

The left side is the equation for a circle with center x = -2, y = 0 and the right side is the equation for a circle with center x = 0 and y = -2.

So the set is the circle described on the left side with radius less than the circle on the right side.
Is this correct?
And if I want to sketch the set?

No that isn't correct. Simplify it and look at the inequality you get and you will see how to graph it.

 

[MaxManus]

Thanks for the help
x + iy
|z+2| < |z+2i|
sqrt((x+2)^2 + y^2) < sqrt(x^2 + (y+2)^2)
if a and b are nonegative numbers and a < b, then a2<b2
(x+2)^2 + y^2 < x^2 + (y+2)^2
x^2 + 4x +x + y^2 < x^2 + y^2 +4y + 4
4x < 4y
x < y

sketch:
divide the x,iy plate with x = y. The set is on the left side

 

[LCKurtz]

Good. Hopefully that fits your intuition about what points are closer to -2 than to -2i.