document.write( "Question 462738: Determine (-2+i3)^5 in the polar form and in cartesian form \n" ); document.write( "
Algebra.Com's Answer #317167 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
Determine (-2+i3)^5 in the polar form and in cartesian form
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document.write( "The other tutor got the angle in the 4th quadrant instead\r\n" );
document.write( "of the 2nd quadrant, and also forgot to multiply the angle\r\n" );
document.write( "by 5, which also would have been wrong since he got the\r\n" );
document.write( "angle in the wrong quadrant.\r\n" );
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document.write( "Draw the vector x + iy = -2 + i3 which is the \r\n" );
document.write( "vector from (0,0) to (-2,3), and let its length \r\n" );
document.write( "(magnitude, modulus, absolute value) be r, and its\r\n" );
document.write( "angle (argument) be q:\r\n" );
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document.write( "Then we draw a perpendicular (in green) from the\r\n" );
document.write( "tip of the vector to the x-axis:\r\n" );
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document.write( "x + iy = eiq = r⋅cis(q) = r⋅(cosq + i⋅sinq)\r\n" );
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document.write( "We calculate r and q)\r\n" );
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document.write( "r² = x² + y²\r\n" );
document.write( "r² = (-2)² + (3)²\r\n" );
document.write( "r² = 4 + 9\r\n" );
document.write( "r² = 13\r\n" );
document.write( " r = √13\r\n" );
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document.write( "tanq = y/x\r\n" );
document.write( "tanq = (3)/(-2) = -1.5\r\n" );
document.write( "q = 123.69° = 2.16 radians\r\n" );
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document.write( "We use:\r\n" );
document.write( "(x + iy)n = rnei·nq = rncis(nq) = rn[cos(nq) + i·sin(nq)]\r\n" );
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document.write( "with n = 5, q = 123.69°, r = √13\r\n" );
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document.write( "(√13)5 = 609.3381656\r\n" );
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document.write( "5q = 5(123.69°) = 618.45° which\r\n" );
document.write( "is coterminal with 258.45°\r\n" );
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document.write( "609.3381656[cos(258.45°) + i·sin(258.45°)] =  -122-i·597\r\n" );
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document.write( "Edwin

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