document.write( "Question 181601: Okay this is my last question.
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\n" ); document.write( "\n" ); document.write( "14-2i Over 3+i.
\n" ); document.write( "I dont know if that makes sense to you\r
\n" ); document.write( "\n" ); document.write( "But i think i learned something about taking the conjucate of 3+i which is 3-1, but i dont know where to go from there, or if im even doing it right.
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Algebra.Com's Answer #136206 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
\"%28%2814-2i%29%29%2F%28%283%2Bi%29%29\"
\n" ); document.write( "you multiply it by the conjugate of the denominator over itself
\n" ); document.write( "\"%28%2814-2i%29%29%2F%28%283%2Bi%29%29\" * \"%28%283-i%29%29%2F%28%283-i%29%29\" = \"%28%2842-14i-6i%2B2i%5E2%29%29%2F%28%289%2B3i-3i-i%5E2%29%29\" = \"%28%2842-20i%2B2i%5E2%29%29%2F%28%289-i%5E2%29%29\"
\n" ); document.write( "i^2 = -1, therefore
\n" ); document.write( "\"%28%2842-20i%2B2%28-1%29%29%29%2F%28%289-%28-1%29%29%29\" = \"%28%2842-20i-2%29%29%2F%28%289%2B1%29%29\" = \"%28%2840-20i%29%29%2F%2810%29\"
\n" ); document.write( "Divide 10 into both terms
\n" ); document.write( "4 - 2i
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