It can be done in the same manner as the previous

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Unformatted text preview: umerator and denominator by the conjugate of the denominator we will be able to eliminate the i from the denominator. Now that we’ve figured out how to do these let’s go ahead and work the problems. (a) ( 3 - i ) ( 2 - 7i) = 6 - 23i + 7i 2 = -1 - 23i = - 1 - 23 i 3-i = 2 + 7i ( 2 + 7i) ( 2 - 7i) 22 + 7 2 53 53 53 Notice that to officially put the answer in standard form we broke up the fraction into the real and imaginary parts. [Return to Problems] 3 3 ( 9 + i ) 27 + 3i 27 3 (b) = = = +i 9 - i ( 9 - i ) ( 9 + i ) 92 + 12 82 82 [Return to Problems] (c) 8i 8i (1 - 2i ) 8i - 16i 2 16 + 8i 16 8 = = = = +i 1 + 2i (1 + 2i ) (1 - 2i ) 12 + 22 5 55 [Return to Problems] © 2007 Paul Dawkins 55 http://tutorial.math.lamar.edu/terms.aspx College Algebra (d) This one is a little different from the previous ones since the denominator is a pure imaginary number. It can be done in the same manner as the previous ones, but there is a slightly easier way to do the problem. First, break up the fraction as follows. 6 - 9i 6 9i...
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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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