June 8th, 2019, 02:12 AM  #1 
Newbie Joined: Jun 2019 From: Ukraine Posts: 6 Thanks: 0  complex plane
Hello to all. I am new and I need your help. Task: Build an area on the complex plane   z + 1  >  1  z  Last edited by skipjack; June 8th, 2019 at 03:25 AM. 
June 8th, 2019, 03:31 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2216 
Let P be the point in the complex plane corresponding to z. The inequality holds if and only if P is closer to the point (1, 0) than to the point (1, 0). P is equidistant from (1, 0) and (0, 1) if and only if P lies on the imaginary axis. Can you proceed from there? 
June 8th, 2019, 03:39 AM  #3 
Newbie Joined: Jun 2019 From: Ukraine Posts: 6 Thanks: 0 
I don't quite understand how to plot

June 8th, 2019, 05:03 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2216 
How to plot what?

June 8th, 2019, 05:24 AM  #5 
Newbie Joined: Jun 2019 From: Ukraine Posts: 6 Thanks: 0 
how to build a graph?

June 8th, 2019, 05:38 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2216 
I wasn't sure what "build an area" meant. The inequality is satisfied if and only if Re(z) > 0.

June 8th, 2019, 05:49 AM  #7 
Newbie Joined: Jun 2019 From: Ukraine Posts: 6 Thanks: 0 
As I understood you need to build a graph in the complex plane of this inequality.

June 8th, 2019, 05:57 AM  #8 
Newbie Joined: Jun 2019 From: Ukraine Posts: 6 Thanks: 0 
I think this is correct but how to do it


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