1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex analysis, graph inequality

  1. Sep 11, 2011 #1
    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. jcsd
  3. Sep 11, 2011 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  4. Sep 12, 2011 #3
    I now see why |x|>2 is NOT a circle.

    I got it confused with this formula |z-z[itex]_{0}[/itex]|=[itex]\rho[/itex]

    Because |z|>2 is a circle.

    And I still have a hard time seeing why |z|>2 is a circle.
     
  5. Sep 12, 2011 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    |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?
     
  6. Sep 12, 2011 #5
    Yes, I see now.

    Thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Complex analysis, graph inequality
Loading...