Stitz-Zeager_College_Algebra_e-book

Once t 2 the orientation reverses and we start to head

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: teger and Arg(z ) = π . 2 2 4. As in the previous problem, we write z = −117 = −117 + 0i so Re(z ) = −117 and Im(z ) = 0. The number z = −117 corresponds to the point (−117, 0), and this is another instance where we can determine the polar form ‘by eye.’ The point (−117, 0) is 117 units away from the origin along the negative x-axis. Hence, r = |z | = 117 and θ = π + 2π = (2k + 1)πk for integers k . We have arg(z ) = {(2k + 1)π : k is an integers}. Only one of these values, θ = π , just barely lies in the interval (−π, π ] which means and Arg(z ) = π . We plot z along with the other numbers in this example below. Imaginary Axis z = −2 + 4i 4i 3i z = 3i 2i i z = −117 −117 −2 −1 −i 1 2 34 √ z = 3−i Real Axis Now that we’ve had some practice computing the modulus and argument of some complex numbers, it is time to explore their properties. We have the following theorem. Theorem 11.14. Properties of the Modulus: Let z and w be complex numbers. • |z | is the distance from z to 0 in the complex...
View Full Document

This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

Ask a homework question - tutors are online