document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=616169>Question 616169</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you write (2 square root of 3) - 2i in polar notation?  I keep getting 4(cos 150 + isin 150).  However, the answer key in my book says it's 4(cos 330 + isin 330).  Help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #387527 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=387527\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> How do you write (2 square root of 3) - 2i in polar notation? I keep getting 4(cos 150 + isin 150). However, the answer key in my book says it's 4(cos 330 + isin 330). Help.<BR>\n" );
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document.write( "2*sqrt(3) - 2i corresponds to (2sqrt(3),-2) which is in the 4th Quadrant.<BR>\n" );
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document.write( "r = sqrt[(2sqrt(3))^2 + 2^2] = sqrt[16] = 4<BR>\n" );
document.write( "theta = tan^-1(-2/(2sqrt(3)) = tan^-1(-1/sqrt(3)) = -30 degrees or +330<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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