Thread: help with complex variables

I was given the problem:

Sketch the following sets and determine which are domains:

(a) z-2+i < 1
(b) 2z+3 > 3
(c) Im(z) > 1
(d) Im(z) = 1
(e) 0 < or equal to arg(z) > or equal to pi/4 and z not equal to 0
(f) z-f > or equal to z

I have no idea how to start this. I know you cannot show sketches on this, but can anyone give me ideas of what I am doing for this question?

Originally Posted by LCopper2010

I was given the problem:

Sketch the following sets and determine which are domains:

(a) z-2+i < 1
(b) 2z+3 > 3
(c) Im(z) > 1
(d) Im(z) = 1
(e) 0 < or equal to arg(z) > or equal to pi/4 and z not equal to 0
(f) z-f > or equal to z

I have no idea how to start this. I know you cannot show sketches on this, but can anyone give me ideas of what I am doing for this question?

If that is true, then why have you been asked to do these?

Originally Posted by LCopper2010

I was given the problem:

Sketch the following sets and determine which are domains:

(a) z-2+i < 1
(b) 2z+3 > 3
(c) Im(z) > 1
(d) Im(z) = 1
(e) 0 < or equal to arg(z) > or equal to pi/4 and z not equal to 0
(f) z-f > or equal to z

I have no idea how to start this. I know you cannot show sketches on this, but can anyone give me ideas of what I am doing for this question?

|z- a| is the distance form point z to point a in the complex plane. |z- 2+i|= |z- (2-i)| is the distance from the point 2-i to z. The set of all z such that that distance is equal to 1 is the circle with center at 2- i and radius 1. What does that tell you about |z-(2-i)|< 1 and |z-(2-i)|> 1?

The imaginary part of z, Im(z), is the "y" coordinate on the complex plane so Im(z)= 1 is a horizontal line.