How would you draw this? (1 Viewer)

[Dubble-u25]

{z e C | |z − i| < |z − 1| }??

Thankyou

 

[nightweaver066]

{z e C | |z − i| < |z − 1| }??

Thankyou

Oh wow. Similar to the one that Carrotsticks posted earlier today haha.

|z - i| starts from 1 on the imaginary axis.

|z - 1| starts from 1 on the real axis.

Perpendicular bisector of the chord joining those two points is y = x

For |z - i| <= |z - 1|, has to be the region below and including y = x, therefore the region of

 

[Dubble-u25]

Really? haha sorry
hmmm... how does |z- i| and |z - 1| both look like? are they still circles?

 

[Carrotsticks]

Really? haha sorry
hmmm... how does |z- i| and |z - 1| both look like? are they still circles?

Not quite.

The locus is no longer |z-i| = k or |z-1| = k, which would give you a circle.

Now, the two moduli are thrown in together with an inequality in between them, which yields a totally different locus (or set of points satisfying the condition).

 

[nightweaver066]

Wait.. do you do MX2 (your signature)?

|z| means the displacement of the complex number, z, from the origin.

Similarly, by shifting it upwards 1 unit, |z - i|, means the displacement of the complex number, z, from 1 unit above the origin, i.e. (0, 1).

Same thing with |z - 1|, just meaning the displacement of the complex number, z, from 1 unit to the right of the origin, i.e. (1, 0).

First look for the region where equality holds, i.e. the displacement from |z - 1| and |z - i| are equal. You find that this is y = x (the perpendicular bisector of the chord between (1, 0) and (0, 1) ).

Then you determine whether it will be above or below the line by testing a point seeing if it satisfies the condition.

 

[Dubble-u25]

ahh i see... thanks for clearing it up