document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=6751>Question 6751</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how do u solve problems like these:\r<BR>\n" );
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document.write( "i^29\r<BR>\n" );
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document.write( "i^155 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #3723 by </B><A HREF=/tutors/aboutme.mpl?userid=prince_abubu><B>prince_abubu(198)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=prince_abubu><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=3723\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I take it that i is the imaginary number that is equivalent to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+sqrt%28-1%29+ ALIGN=MIDDLE  ALT=\"+sqrt%28-1%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "There's a trick here that many teachers don't show when computing <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+i%5En+ ALIGN=MIDDLE  ALT=\"+i%5En+\"   DISABLED_style= \"max-width:90%; height:auto;\" >. We're gonna have to start from scratch here.\r<BR>\n" );
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document.write( "i^1 = i<BR>\n" );
document.write( "i^2 = i * i = -1<BR>\n" );
document.write( "i^3 = i^2 * i = -i<BR>\n" );
document.write( "i^4 = i^3 * i = -i * i = 1\r<BR>\n" );
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document.write( "i^5 = i^4 * i = i<BR>\n" );
document.write( "i^6 = i^5 * i = i * i = -1<BR>\n" );
document.write( "i^7 = i^6 * i = -1 * i = -i<BR>\n" );
document.write( "i^8 = i^7 * i = -i * i = 1\r<BR>\n" );
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document.write( "and so on.\r<BR>\n" );
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document.write( "If you noticed, it goes from i ---> -1 ----> -i ----> 1 and back again to i. It's like a cycle. Another thing that you probably noticed is that every power of i divisible by 4 equals 1, so i^4 = i^8 = ...= i^40 = ... = i^400 = 1.\r<BR>\n" );
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document.write( "What about the rest? Actually all you have to do is to take that power n, divide by 4, and pay attention to the remainder.\r<BR>\n" );
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document.write( "If the remainder is 0, then i^n = 1<BR>\n" );
document.write( "If the remainder is 1, then i^n = -1<BR>\n" );
document.write( "If the remainder is 2, then i^n = i<BR>\n" );
document.write( "If the remainder is 3, then i^n = -i\r<BR>\n" );
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document.write( "So let's take i^29. n = 29. Divide that by 4. You'll get 7 remainder 1. Just because it's remainder 1, i^29 = -1. I think you can take it from here.  </SPAN>\n" );
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