Return to problems 2007 paul dawkins 54

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Unformatted text preview: form. The standard form for complex numbers does not have an i2 in it. This however is not a problem provided we recall that i 2 = -1 Using this we get, 7i ( -5 + 2i ) = -35i + 14 ( -1) = -14 - 35i We also rearranged the order so that the real part is listed first. [Return to Problems] (b) In this case we will FOIL the two numbers and we’ll need to also remember to get rid of the i2. (1 - 5i ) ( -9 + 2i ) = -9 + 2i + 45i - 10i 2 = -9 + 47i - 10 ( -1) = 1 + 47i [Return to Problems] (c) Same thing with this one. ( 4 + i ) ( 2 + 3i ) = 8 + 12i + 2i + 3i 2 = 8 + 14i + 3 ( -1) = 5 + 14i [Return to Problems] (d) Here’s one final multiplication that will lead us into the next topic. (1 - 8i ) (1 + 8i ) = 1 + 8i - 8i - 64i 2 = 1 + 64 = 65 Don’t get excited about it when the product of two complex numbers is a real number. That can and will happen on occasion. [Return to Problems] © 2007 Paul Dawkins 54 http://tutorial.math.lamar.edu/terms.aspx College Algebra In the final part of the previous example we multip...
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