Alg_Complete

7i 5 2i 35i 14i 2 now this is where the small

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Unformatted text preview: hat we’re thinking of that term as a complex number and in © 2007 Paul Dawkins 53 http://tutorial.math.lamar.edu/terms.aspx College Algebra general aren’t used. (a) ( -4 + 7i ) + ( 5 - 10i ) = 1 - 3i (b) ( 4 + 12i ) - ( 3 - 15i ) = 4 + 12i - 3 + 15i = 1 + 27i (c) 5i - ( -9 + i ) = 5i + 9 - i = 9 + 4i Next let’s take a look at multiplication. Again, with one small difference, it’s probably easiest to just think of the complex numbers as polynomials so multiply them out as you would polynomials. The one difference will come in the final step as we’ll see. Example 2 Multiply each of the following and write the answers in standard form. (a) 7i ( -5 + 2i ) [Solution] (b) (1 - 5i ) ( -9 + 2i ) [Solution] (c) ( 4 + i ) ( 2 + 3i ) [Solution] (d) (1 - 8i ) (1 + 8i ) [Solution] Solution (a) So all that we need to do is distribute the 7i through the parenthesis. 7i ( -5 + 2i ) = -35i + 14i 2 Now, this is where the small difference mentioned earlier comes into play. This number is NOT in standard...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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