document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=30908>Question 30908</A>: <I><SPAN STYLE=\"word-break: break-all;\"> perform the indicated operation and simplify:   \r<BR>\n" );
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document.write( "                       1+3i/1-2i </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #20059 by </B><A HREF=/tutors/aboutme.mpl?userid=kietra><B>kietra(57)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kietra><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=20059\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> (1+3i)/(1-2i)<BR>\n" );
document.write( "In imaginary numbers, there's a rule that you can never have an i in the denominator.  To get rid of that, you multiply by the inverse of (1-2i) which is (1+2i) in both the denominator and numerator.  You must use FOIL to multiply the numbers. <BR>\n" );
document.write( "(1+3i)/(1-2i)* (1+2i)/(1+2i).  Multiply numerators together and denominators. <BR>\n" );
document.write( "(1+3i)(1+2i)/(1-2i)(1+2i)<BR>\n" );
document.write( "(1 + 2i + 3i + 6i^2)/(1 -2i + 2i - 4i^2) <BR>\n" );
document.write( "(1 + 5i + 6(-1))/(1 -4(-1))<BR>\n" );
document.write( "(1-6 + 5i)/(1 + 4)<BR>\n" );
document.write( "(-5 + 5i)/(5) =now reduce<BR>\n" );
document.write( "5(-1 + i)/5<BR>\n" );
document.write( "-1 + i <BR>\n" );
document.write( " </SPAN>\n" );
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