# Complex analysis, graph inequality

1. Sep 11, 2011

### Fellowroot

1. The problem statement, all variables and given/known data

Sketch the graph

|Re(z)|>2

2. Relevant equations

z=x+iy

3. The attempt at a solution

|Re(z)|>2

|Re(x+iy)|>2

|x|>2

|x-0|>2, this is a circle centered at zero with radius 2

4. My question

What I'm having a hard time with is the | | notation.
Is this the absolute value, or modulus, or something else.

Wolfram Alpha says its an absolute value graph (that I understand) but my books says this is the form of a circle. Which is it?
http://www.wolframalpha.com/input/?i=|Re(z)|>2

I graphed a circle on my paper r=2 at (0,0) but I don't know which part to shade, the inside of the circle or the outside? How can I plug in numbers to test True and False? I know how to graph inequalities in real but complex I'm having trouble with.

Thanks.

2. Sep 11, 2011

### Dick

I have no idea what Wolfram Alpha is thinking about. Ignore it, please? || is modulus or absolute value. They are really the same thing. |x|>2 is perfectly correct. But it's not a circle at all. x=2 is a line, x=(-2) is another line. Those are the boundaries of your region, yes?

3. Sep 12, 2011

### Fellowroot

I now see why |x|>2 is NOT a circle.

I got it confused with this formula |z-z$_{0}$|=$\rho$

Because |z|>2 is a circle.

And I still have a hard time seeing why |z|>2 is a circle.

4. Sep 12, 2011

### Dick

|z|=|x+iy|=sqrt(x^2+y^2). Do you know that? sqrt(x^2+y^2)=2 is the same as x^2+y^2=4. Do you see why x^2+y^2=4 is a circle?

5. Sep 12, 2011

### Fellowroot

Yes, I see now.

Thank you.