document.write( "Question 146485: Invisible numbers...Perform indicated operation. Write results in the form of a+bi\r
\n" ); document.write( "\n" ); document.write( "6 / 5-2i\r
\n" ); document.write( "\n" ); document.write( "Six divided by five minus 2i. \r
\n" ); document.write( "\n" ); document.write( "I started this problem by mutiplying the numerator and the denominator by five minus 2i. I obtained the answer of 30 -12i for the numerator...and I do not know what to do after that.\r
\n" ); document.write( "\n" ); document.write( "Thanks! \n" ); document.write( "
Algebra.Com's Answer #106916 by nerdybill(7384)\"\" \"About 
You can put this solution on YOUR website!
The whole idea is to get rid of any i's you see in the denominator. How do we do this?
\n" ); document.write( "Since we know that i*i = -1
\n" ); document.write( "We would like to apply this bit of info.
\n" ); document.write( "We need to find something to multiply the denominator such that we can rid ourselves of that i.
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\n" ); document.write( "So, if you started with:
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\n" ); document.write( "6/(5-2i)
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\n" ); document.write( "Multiply numerator and denominator with (5+2i)
\n" ); document.write( "6/(5-2i) * (5+2i)/(5+2i)
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\n" ); document.write( "Looking at the denominator, we have:
\n" ); document.write( " (5-2i)(5+2i)
\n" ); document.write( "Applying \"FOIL\", we get (25 + 10i - 10i + 4)
\n" ); document.write( "notice the i's go away leaving just 29.
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\n" ); document.write( "So, let's multiply out:
\n" ); document.write( "6/(5-2i) * (5+2i)/(5+2i)
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\n" ); document.write( "we get:
\n" ); document.write( "6(5+2i)/29
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\n" ); document.write( "(30+12i)/29 \r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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