Find the set of points that satisfy:|z|^2 + |z - 2*i|^2 =< 10

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views
Makadamij
Messages
7
Reaction score
2
Homework Statement
Find and draw the set of all points that satisfy the following condition: |z|^2 + |z - 2*i|^2 =< 10, where z is a complex number.
Relevant Equations
|z| = sqrt(a^2 + b^2)
z = a + b*i
Hello everyone,

I've been struggling quite a bit with this problem, since I'm not sure how to approach it correctly. The inequality form reminds me of the equation of a circle (x^2 + y^2 = r^2), but I have no idea how to be sure about it. Would it help just to simplify the inequality in terms of solving the absolute value of z and z-2i ?

I would be very thankful for any advices and solutions.

Best wishes,

Makadamij
 
Physics news on Phys.org
Hello, thank you for participating in my problem. By expressing and expanding I get:

|x+yi|^2 + |x+(y-2)i|=< 10
x^2 + y^2 + x^2 + (y-2)^2 =< 10
2x^2 + 2y^2 - 4y + 4 =<10
2x^2 + 2y^2 - 4y =< 6 /:2
x^2 + y^2 - 2y =< 3

This can be transformed to the equation of a circle

x^2 + (y-1)^2 - 2 =< 3
x^2 + (y-1)^2 =< 5

So I get the formula of a cirlce with radius sqrt(5), which centre point is at S(0,1).
Are my calculations therefore correct or am I missing something?
 
Makadamij said:
Hello, thank you for participating in my problem. By expressing and expanding I get:

|x+yi|^2 + |x+(y-2)i|=< 10
x^2 + y^2 + x^2 + (y-2)^2 =< 10
2x^2 + 2y^2 - 4y + 4 =<10
2x^2 + 2y^2 - 4y =< 6 /:2
x^2 + y^2 - 2y =< 3

This can be transformed to the equation of a circle

x^2 + (y-1)^2 - 2 =< 3
x^2 + (y-1)^2 =< 5

So I get the formula of a cirlce with radius sqrt(5), which centre point is at S(0,1).
Are my calculations therefore correct or am I missing something?
You made a small mistake near the end.

You could check this yourself by setting ##w = z - i##. Now that you think it's a circle centred at ##0 + i##. You should get ##|w| \le R##.
 
I've spotted the mistake, thank you. The radius is supposed to be 2, not sqrt(5). And also a kind thank you for all the advices. Problem is now solved :)
 
  • Like
Likes   Reactions: scottdave and PeroK