document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=882013>Question 882013</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Rewrite the numerator and denominator in polar or exponential form before carrying out the division.\r<BR>\n" );
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document.write( "(sqrt(3)-i)/(2i+2)\r<BR>\n" );
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document.write( "I know how to approach this problem but I just can't seem to get the same answer as in the textbook. This is my working:\r<BR>\n" );
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document.write( "(sqrt(3)-i) >>> Modulus = 2, Arg = -arctan(1/sqrt(3))= - pi/6\r<BR>\n" );
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document.write( "(sqrt(3)-i)= 2(cos(-pi/6) + sin(-pi/6)i)\r<BR>\n" );
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document.write( "(2i+2) >>> Modulus = 2sqrt(2), Arg = arctan(1) = pi/4\r<BR>\n" );
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document.write( "(2i+2) = 2sqrt(2)(cos(pi/4) + sin(pi/4)i)\r<BR>\n" );
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document.write( "(sqrt(3)-i)/(2i+2) = 2/(2sqrt(2))*e^(-pi/6 - pi/4) = 1/sqrt(2) * e^(-5pi/12)\r<BR>\n" );
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document.write( "The textbook answer is sqrt(2) * e^(-5pi/12). Who is correct? Thank you.\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #532665 by </B><A HREF=/tutors/aboutme.mpl?userid=pj66><B>pj66(7)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=pj66><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=532665\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I don't really understand the e to the power -pi/6 -pi/4 part, so, I solve in standard form(without e to the power).<BR>\n" );
document.write( "All that you have solved is correct. I think the problem is in the division part. <BR>\n" );
document.write( "After putting the values, we get 2*((cos pi/6)-(sin pi/6)i)/(2^(3/2)((cos pi/4)+(sin pi/4)i))<BR>\n" );
document.write( "where -cos pi/6=cos pi/6<BR>\n" );
document.write( "multiply and divide by (cos pi/4)-(sin pi/4)i)<BR>\n" );
document.write( "=> ((cos pi/6 - sin pi/4i)*(cos pi/4 - sin pi/4i))/(sqrt2(cos pi/4 + sinpi/4i)*(cos pi/4 - sin pi/4i)) <BR>\n" );
document.write( "=>((cos pi/6)*(cos pi/4)-(sin pi/6)*(cos  pi/4)i-(cos pi/6)*(sin pi/4)i-(sin pi/6)*(sin pi/4))/(sqrt 2)<BR>\n" );
document.write( "Using cos(x+y)=cosx*cosy-siny*cosx<BR>\n" );
document.write( "and sin(x+y)=sinx*cosy-siny*cosx<BR>\n" );
document.write( "=>(2^(-1/2))*(cos(pi/6 + pi/4)-sin(pi/6 + pi/4)i)<BR>\n" );
document.write( "I hope this helps. </SPAN>\n" );
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