Stitz-Zeager_College_Algebra_e-book

Starting with t ln1 0 we get5 2 1 for t ln2 we

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Unformatted text preview: in Quadrant IV, θ is a Quadrant IV angle. We know tan(θ) = √1 = − 33 , so θ = − π + 2πk 6 3 for integers k . Hence, arg(z ) = − π + 2πk : k is an integer . Of these values, only θ = − π 6 6 satisfies the requirement that −π < θ ≤ π , hence Arg(z ) = − π . 6 2. The complex number z = −2 + 4i has Re(z ) = −2, Im(z ) = 4, and is associated with the point P (−2, 4). Our next task is to find a polar representation (r,√) for P where r ≥ 0. θ √ Running through the usual calculations gives r = 2 5, so |z | = 2 5. To find θ, we get tan(θ) = −2, and since r > 0 and P lies in Quadrant II, we know θ is a Quadrant II angle. Using a reference angle approach,6 we find θ = π − arctan(2) + 2πk for integers k . Hence arg(z ) = {π − arctan(2) + 2πk : k is an integer}. Only θ = π − arctan(2) satisfies the requirement −π < θ ≤ π , so Arg(z ) = π − arctan(2). 2 In case you’re wondering, the use of the absolute value notation |z | for modulus will be explained shortly. Note that since arg(z...
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