document.write( "Question 364688: If i^2 =-1, then i^142 =? \n" ); document.write( "

Algebra.Com's Answer #260038 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
Work out the pattern,
\n" ); document.write( " $\"i%5E0=1\"$
\n" ); document.write( " $\"i%5E1=i\"$
\n" ); document.write( " $\"i%5E2=-1\"$
\n" ); document.write( " $\"i%5E3=-i\"$
\n" ); document.write( " $\"i%5E4=1\"$
\n" ); document.write( "You see the pattern repeats every $\"4\"$ .
\n" ); document.write( "So is $\"142\"$ an integer multiple of $\"0%2B4n\"$ , $\"1%2B4n\"$ , $\"2%2B4n\"$ , or $\"3%2B4n\"$ ???
\n" ); document.write( "Since it's even it can't be either odd combination.
\n" ); document.write( " $\"0%2B4n=142\"$
\n" ); document.write( " $\"4n=142\"$
\n" ); document.write( " $\"2n=71\"$
\n" ); document.write( "Nope that's not it, so it must be
\n" ); document.write( " $\"2%2B4n=142\"$
\n" ); document.write( " $\"4n=140\"$
\n" ); document.write( " $\"n=35\"$
\n" ); document.write( "So then,
\n" ); document.write( " $\"highlight%28i%5E2=i%5E142=-1%29\"$
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