Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook