Alg_Complete

# We next need to address an issue on dealing with

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Unformatted text preview: 3 9 = - =2i 2i 2i i 2 Now, we want the i out of the denominator and since there is only an i in the denominator of the first term we will simply multiply the numerator and denominator of the first term by an i. 6 - 9i 3 ( i ) 9 3i 9 3i 9 9 = - = 2- = - = - - 3i 2i i (i) 2 i 2 -1 2 2 [Return to Problems] The next topic that we want to discuss here is powers of i. Let’s just take a look at what happens when we start looking at various powers of i. i1 = i i1 = i i 2 = -1 i 2 = -1 i 3 = i × i 2 = -i i 3 = -i i 4 = ( i 2 ) = ( -1) = 1 i4 = 1 i5 = i × i4 = i i5 = i i 6 = i 2 × i 4 = ( -1)(1) = -1 i 6 = -1 i 7 = i × i 6 = -i i 7 = -i 2 2 i 8 = ( i 4 ) = (1) = 1 2 2 i8 = 1 Can you see the pattern? All powers if i can be reduced down to one of four possible answers and they repeat every four powers. This can be a convenient fact to remember. We next need to address an issue on dealing with square roots of negative numbers. From the section on radicals we know that we can do the following....
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