document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1208825>Question 1208825</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Write each expression in the standard form a + bi.\r<BR>\n" );
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document.write( "1. i^(-23)\r<BR>\n" );
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document.write( "2. (1 + i^3\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #847295 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=847295\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "I'll focus on problem #1 only.<BR>\n" );
document.write( "Problem #2 seems to have a typo in it. I have a feeling it's supposed to say (1+i)^3, but I'm not sure.\r<BR>\n" );
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document.write( "Anyway, with problem #1, let's consider i^23 and worry about the negative exponent later.<BR>\n" );
document.write( "To evaluate i^23, we divide the exponent by 4 and look at the remainder.<BR>\n" );
document.write( "Refer to this <a  rel=nofollow HREF=\"https://www.algebra.com/algebra/homework/complex/Complex_Numbers.faq.question.1207482.html\">similar question</a> to see why we divide by 4.\r<BR>\n" );
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document.write( "23/4 = 5 <font color=blue>remainder 3</font><BR>\n" );
document.write( "This means, i^23 = i^<font color=blue>3</font> = -i\r<BR>\n" );
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document.write( "Another way we can arrive at that is to say:<BR>\n" );
document.write( "i^(23) = i^(20+3)<BR>\n" );
document.write( "= i^(20)*i^3<BR>\n" );
document.write( "= i^(5*4)*i^3<BR>\n" );
document.write( "= (i^4)^5*i^3<BR>\n" );
document.write( "= (1)^5*i^3<BR>\n" );
document.write( "= i^3<BR>\n" );
document.write( "= -i<BR>\n" );
document.write( "Notice I rewrote 23 as 20+3. The 20 is the largest multiple of 4 just short of 23.<BR>\n" );
document.write( "The useful exponent rules are a^b*a^c = a^(b+c) and (a^b)^c = a^(b*c).\r<BR>\n" );
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document.write( "There are probably other ways of determining that i^23 = i^3 = -i.\r<BR>\n" );
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document.write( "Now to account for the negative exponent, we could have these steps<BR>\n" );
document.write( "i^(-23) = 1/(i^23)<BR>\n" );
document.write( "= 1/(-i)<BR>\n" );
document.write( "= i/(-i^2)<BR>\n" );
document.write( "= i/(-(-1))<BR>\n" );
document.write( "= <font color=red>i</font><BR>\n" );
document.write( "The jump from the 2nd line to the 3rd line is when we multiply top/bottom by i. <BR>\n" );
document.write( "Doing this step will convert the imaginary denominator to a real number.\r<BR>\n" );
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document.write( "Answer: i^(-23) = <font color=red size=4>i</font><BR>\n" );
document.write( "Verification with <a  rel=nofollow HREF=\"https://www.wolframalpha.com/input?i=i%5E%28-23%29\">WolframAlpha</a><BR>\n" );
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