Stitz-Zeager_College_Algebra_e-book

I 4 4 34 complex zeros and the fundamental theorem of

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Unformatted text preview: ial Functions 1. (1 − 2i) − (3 + 4i) 4. 2. (1 − 2i)(3 + 4i) 5. 3. 1 − 2i 3 − 4i √ √ −3 −12 (−3)(−12) 6. (x − [1 + 2i])(x − [1 − 2i]) Solution. 1. As mentioned earlier, we treat expressions involving i as we would any other radical. We combine like terms to get (1 − 2i) − (3 + 4i) = 1 − 2i − 3 − 4i = −2 − 6i. 2. Using the distributive property, we get (1 − 2i)(3 + 4i) = (1)(3) + (1)(4i) − (2i)(3) − (2i)(4i) = 3 + 4i − 6i − 8i2 . Recalling i2 = −1, we get 3 + 4i − 6i − 8i2 = 3 − 2i − (−8) = 11 − 2i. 3. How in the world are we supposed to simplify 1−2i ? Well, we deal with the denominator 3−4i 3 − 4i as we would any other denominator containing a radical, and multiply both numerator and denominator by 3 + 4i (the conjugate of 3 − 4i).5 Doing so produces 1 − 2i 3 + 4 i (1 − 2i)(3 + 4i) 11 − 2i 11 2 · = = = − i 3 − 4i 3 + 4 i (3 − 4i)(3 + 4i) 25 25 25 4. We use property 2 of Definition 3.4 first, then √ apply the rules of radicals applicable to real √ √ √ √√ 2 3 · 12 = − 36 = −6. radicals to get −3 −12 = i 3 i 12 = i 5. We adhere to the order of operations here and perform the multiplication before the radical √ to get (−3)(−12) = 36 = 6. 6. We can brute...
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