document.write( "Question 1074317: How to find a three cube root of -64i? \n" ); document.write( "
Algebra.Com's Answer #689015 by math_helper(2461)\"\" \"About 
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\"+y+=+-64i+\"\r
\n" ); document.write( "\n" ); document.write( "Using Euler's:
\n" ); document.write( " y = -64*e^((i*(pi)/2)) \r
\n" ); document.write( "\n" ); document.write( "y^(1/3) = (-64*e^((i*(pi)/2)) ) ^(1/3)
\n" ); document.write( "y^(1/3) = -4* ( e^(i*(pi)/2) ) ^(1/3) \r
\n" ); document.write( "\n" ); document.write( "Now, before multiplying through by 1/3 on the right notice that because e^(i*pi*a) = cos(pi*a) + i*sin(pi*a) the value repeats every 2k*pi, so insert a 2k*pi into the exponent:\r
\n" ); document.write( "\n" ); document.write( " y^(1/3) = -4[e^(i*(pi)/2 + 2k(pi)) ]^(1/3) [ eq 1 ]\r
\n" ); document.write( "\n" ); document.write( "Let the roots be r0, r1, r2 (corresponding to k=0,1,2):\r
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\n" ); document.write( "\n" ); document.write( " k=0 —> \"+r0+=+-4%28cos%28pi%2F6%29+%2B+i%2Asin%28pi%2F6%29%29+\"
\n" ); document.write( " \"+r0+=+-4+%28sqrt%283%29%2F2+%2B+i%2A%281%2F2%29%29+=+highlight%28-2%2Asqrt%283%29+-+2i%29+\"\r
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\n" ); document.write( "\n" ); document.write( "I will leave it to you to find the other two roots (set k=1 in [eq 1] and simplify , then set k=2 and simplify )
\n" ); document.write( "Let me know if you have any questions.
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