document.write( "Question 130377: Use De Moivre's Formula to find (-1+2i)^36 \n" ); document.write( "
Algebra.Com's Answer #95228 by Edwin McCravy(20086)\"\" \"About 
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Use De Moivre's Formula to find (-1+2i)^36
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document.write( "To find (x+yi)n\r\n" );
document.write( "First change x+yi to trigonometric form.\r\n" );
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document.write( "To change x + iy to trigonometric form, we plot\r\n" );
document.write( "the point (x,y), and draw a right triangle with\r\n" );
document.write( "vertices (x,y), (x,0), and (0,0)\r\n" );
document.write( "We calculate the hypotenuse r from r² = x² + y²,\r\n" );
document.write( "which is called the modulus or absolute value\r\n" );
document.write( "of the complex number x+yi\r\n" );
document.write( "We calculate the angle q\r\n" );
document.write( "from either\r\n" );
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document.write( "tanq = \"y%2Fx\",\r\n" );
document.write( "or \r\n" );
document.write( "sinq = \"y%2Fr\",\r\n" );
document.write( "or\r\n" );
document.write( "cosq = \"x%2Fr\",\r\n" );
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document.write( "Then x + yi has the trigonometric form:\r\n" );
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document.write( "x + yi = r(cosq + i·sinq)\r\n" );
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document.write( "Then we use the formula:\r\n" );
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document.write( "[r(cosq + i·sinq)]n = rn[cos(nq) + i·sin(nq)]\r\n" );
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document.write( "The real part of -1+2i is -1\r\n" );
document.write( "The imaginary part of -1+2i is 2, the coefficient of i.\r\n" );
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document.write( "We plot the point (-1,2), and draw a right triangle with\r\n" );
document.write( "vertices (-1,2), (-1,0), and (0,0)\r\n" );
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document.write( "Next we calculate r:\r\n" );
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document.write( "r² = x² + y²\r\n" );
document.write( "r² = (-1)² + 2²\r\n" );
document.write( "r² = 1 + 4\r\n" );
document.write( "r² = 5\r\n" );
document.write( "r = \"sqrt%285%29\"\r\n" );
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document.write( "Now we need to calculate the angle q\r\n" );
document.write( "indicated by the arc:\r\n" );
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document.write( "We first calculate its reference angle from, say\r\n" );
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document.write( "tan(refq) = \"2%2F1\",\r\n" );
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document.write( "So we find the reference angle from the inverse tangent\r\n" );
document.write( "on the calculator:\r\n" );
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document.write( "refq = 63.4349°\r\n" );
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document.write( "Now we know that q is in \r\n" );
document.write( "Quadrant II, so we subtract the reference angle from \r\n" );
document.write( "180° and get q = 116.5651°\r\n" );
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document.write( "So the trigonometric form is\r\n" );
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document.write( "-1 + 2i = \"sqrt%285%29\"[cos(116.5651°)+ i·sin(116.5651°)]\r\n" );
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document.write( "So using the formula\r\n" );
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document.write( "[r(cosq + i·sinq)]n = rn[cos(nq) + i·sin(nq)]\r\n" );
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document.write( "[\"sqrt%285%29\"(cos116.5651° + i·sin116.5651°)36 =\r\n" );
document.write( "\"%28sqrt%285%29%29%5E36\"[cos(36·116.5651°) + i·sin(36·116.5651°)] = \r\n" );
document.write( "\"%285%5E%281%2F2%29%29%5E36\"[cos(4196.3436°) + i·sin(4196.3436°)] = \r\n" );
document.write( "\"5%5E18\"[cos(4196.3436°) + i·sin(4196.3436°)]\r\n" );
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document.write( "Now we can replace the angle 4196.3436° by its smallest \r\n" );
document.write( "positive coterminal angle.\r\n" );
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document.write( "To do this we divide the angle by 360° and get 11.65651.  The\r\n" );
document.write( "whole part is 11 so we subtract 360° 11 times, which means we \r\n" );
document.write( "subtract 11·360° or 3960° from 4196.3436° and get\r\n" );
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document.write( "4196.3436° - 3960° = 236.3436°\r\n" );
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document.write( "So the answer in trig form is\r\n" );
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document.write( "518(cos236.3436° + i·sin236.3436°)\r\n" );
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document.write( "Using a calculator we get\r\n" );
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document.write( "-2.114 - 3.175i for the answer in standard form.\r\n" );
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document.write( "Edwin

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