Question

Compare and contrast the absolute value of a real number to that of a complex number

Answer 1

Consider a real number in two dimensional plane that X axis and Y axis having ordered pair : P(x,y) or (3,4).

To find it's absolute value we will find it's Distance from origin.

As Coordinate of Origin = (0,0)

Absolute Value of OP = $\left | OP \right |$

     =  $\sqrt{(0-x)^2+(0-y)^2}=\sqrt {x^2+y^2}$

    If the point is Q(3,4)

$\left | OQ \right |=\sqrt{3^2+4^2}=\sqrt{5^2}=5$

In case of a complex number :

Z = a + bi , we consider two axes Real Axis and Imaginary Axis.

We represent Z that is (a,b) in the same way that we represent point in two dimensional plane.

To find it's absolute value , we will find it's distance from origin.

$\left | Z \right |= \sqrt{a^2+b^2}$

If $Z_{1}$ = 6 + 8 i

$\left | Z_{1}\right |=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$

Answer 2

The absolute value of a real number is a positive value of the number. Which means that the absolute value is the distance from zero of the number line. However, that of the complex numbers is the distance from the origin to the point in a complex plane,