document.write( "Question 984942: Find the sixth roots of -64. Enter the solutions in standard form, in any order.
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Algebra.Com's Answer #605794 by ikleyn(52797)\"\" \"About 
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\n" ); document.write( "Notice that     -64 = 64(cos(180°) + i*sin(180°)).\r
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\n" ); document.write( "\n" ); document.write( "1)  2(cos(30°) + i*sin(30°)) = \"2%28sqrt%283%29%2F2+%2B+i%2A%281%2F2%29%29\" = \"sqrt%283%29+%2B+i\";                                                 (Notice that 30° = 180°/6)\r
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\n" ); document.write( "\n" ); document.write( "2)  2(cos(30°+60°) + i*sin(30°+60°)) = 2(cos(90°) + i*sin(90°)) = \"2%2A0+%2B+2%2Ai%29\" = \"2i\";       (Notice that 60° = 360°/6)\r
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\n" ); document.write( "\n" ); document.write( "3)  2(cos(30°+120°) + i*sin(30°+120°)) = 2(cos(150°) + i*sin(150°)) = \"2%2A%28%28-sqrt%283%29%2F2%29+%2B+i%2A%281%2F2%29%29\" = \"-sqrt%283%29+%2B+i%29\";\r
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\n" ); document.write( "\n" ); document.write( "4)  2(cos(30°+180°) + i*sin(30°+180°)) = 2(cos(210°) + i*sin(210°)) = \"2%2A%28%28-sqrt%283%29%2F2%29+%2B+i%2A%28-1%2F2%29%29\" = \"-sqrt%283%29+-+i%29\";\r
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\n" ); document.write( "\n" ); document.write( "5)  2(cos(30°+240°) + i*sin(30°+240°)) = 2(cos(270°) + i*sin(270°)) = \"2%2A%280+-+i%29\" = \"-2i%29\";\r
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\n" ); document.write( "\n" ); document.write( "6)  2(cos(30°+300°) + i*sin(30°+300°)) = 2(cos(330°) + i*sin(330°)) = \"2%2A%28%28sqrt%283%29%2F2%29+%2B+i%2A%28-1%2F2%29%29\" = \"sqrt%283%29+-+i%29\".\r
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\n" ); document.write( "\n" ); document.write( "If you want to see my lessons on complex numbers in this site, look in this \r
\n" ); document.write( "\n" ); document.write( "    - REVIEW of lessons on complex numbers\r
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\n" ); document.write( "\n" ); document.write( "    - How to take a root of a complex number.\r
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