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Stitz-Zeager_College_Algebra_e-book

# Once t 2 the orientation reverses and we start to head

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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...
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