document.write( "Question 54751: Convert the following to polar form, answer in radians
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Algebra.Com's Answer #36958 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
Convert the following to polar form, answer in radians\r\n" );
document.write( "(i) 1 + j3\r\n" );
document.write( "(ii)5 - j2\r\n" );
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document.write( "In general, the picture of a + jb is the vector that \r\n" );
document.write( "connects the origin (0,0) to the point (a,b)  \r\n" );
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document.write( "Therefore, the picture of 1 + j3 is the vector that \r\n" );
document.write( "connects the origin to the point (1,3) is this vector:\r\n" );
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document.write( "Now we draw in the y-coordinate:\r\n" );
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document.write( "That triangle has base 1, and the green side is 3, so the vector is \r\n" );
document.write( "the hypotenuse and that equals \r\n" );
document.write( " _____    ___    __\r\n" );
document.write( "Ö1²+3² = Ö1+9 = Ö10\r\n" );
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document.write( "The angle between the vector and the x-axis has a tangent of 3/1, so\r\n" );
document.write( "the angle is 1.249 radians, approximately.  So the polar form can be \r\n" );
document.write( "written any of the following ways:\r\n" );
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document.write( "Ö10[sin(1.249) + j·cos(1.249)]\r\n" );
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document.write( "Ö10 cjs(1.249)\r\n" );
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document.write( "Ö10 Ð 1.249\r\n" );
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document.write( "(ii)5 - j2\r\n" );
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document.write( "The picture of 5 - j2 is the vector that \r\n" );
document.write( "connects the origin to the point (5,-2) is this vector:\r\n" );
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document.write( "Now we draw in the y-coordinate:\r\n" );
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document.write( "That triangle has top side 5, and the green side is -2, so the vector \r\n" );
document.write( "is the hypotenuse and that equals \r\n" );
document.write( " ________    ____    __\r\n" );
document.write( "Ö5²+(-2)² = Ö25+4 = Ö29\r\n" );
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document.write( "The angle between the vector and the x-axis has a tangent of -2/5, so\r\n" );
document.write( "the angle is -.381 radians, approximately.  So the polar form can be \r\n" );
document.write( "written any of the following ways:\r\n" );
document.write( " __\r\n" );
document.write( "Ö29[sin(-.381) + j·cos(-.381)]\r\n" );
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document.write( "Ö29 cjs(-.381)\r\n" );
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document.write( "Ö29 Ð -.381\r\n" );
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document.write( "Edwin
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