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Homework Help: Representation of properties of Complex Numbers in Argand Diagrams

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Draw an argand diagram to represent the follwing property:
    real(z) < abs(z) < real(z)+img(z)


    2. Relevant equations
    z = x+iy;
    real(z) = x
    abs(z) = sqrt(x^2 + y^2)
    img(z) = y

    3. The attempt at a solution
    substituting original expression with x, y, and sqrt(x^2 + y^2) two inequalities are obtained:
    1. x^2 < x^2 + y^2; which simplifies to y > 0
    2. x^2 + y^2 < x^2 + y^2 + 2*x*y which simplifies to x*y > 0
    now the solution for the first inequality is clear: all the region in an argand diagram above the x-axis(real axis).
    the second inequality remains unclear.
     
  2. jcsd
  3. Mar 10, 2010 #2

    HallsofIvy

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    Science Advisor

    In an "Argand diagram" a complex number, z= x+ iy, is represented as a point (x,y) in the complex plane. |z| is the straight line distance from (0, 0) to (x, y), Re(z)= x is the x component, and Im(z)= y is the y component.

    If you draw the perpendicular form (x, y) to (x, 0), you should see that those three numbers are the lengths of the sides of the right triangle formed by (0, 0), (x, 0), and (x, y).

    The first inequality, Re(z)< |z| just states that the hypotenuse of a right triangle is longer than either leg and can be proved exactly as you say.

    The second inequality, |z|< Re(z)+ Im(z) is the "triangle inequality"- since a straight line is the shortest distance between two points, going directly from (0,0) to (x, y) along the hypotenuse, distance |z|, is shorter than going from (0,0) to (x, 0), distance Re(z), and then from (x, 0) to (x, y), distance Im(z).
     
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