SOLUTION: Convert the following to polar form, answer in radians (i) 1 + j3 (ii)5 - j2

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Question 54751: Convert the following to polar form, answer in radians
(i) 1 + j3
(ii)5 - j2
Answer by Edwin McCravy(6936) About Me  (Show Source):
You can put this solution on YOUR website!
Convert the following to polar form, answer in radians
(i) 1 + j3
(ii)5 - j2

In general, the picture of a + jb is the vector that 
connects the origin (0,0) to the point (a,b)  

Therefore, the picture of 1 + j3 is the vector that 
connects the origin to the point (1,3) is this vector:

+graph%28+200%2C+200%2C+-2%2C+3%2C+-2%2C+3%2C+3%2A%28sqrt%281-x%29%29%2F%28sqrt%281-x%29%29%2Ax%2A%28sqrt%28x%29%29%2F%28sqrt%28x%29%29+%29+

Now we draw in the y-coordinate:

+graph%28+200%2C+200%2C+-2%2C+3%2C+-2%2C+3%2C+3%2A%28sqrt%281-x%29%29%2F%28sqrt%281-x%29%29%2Ax%2A%28sqrt%28x%29%29%2F%28sqrt%28x%29%29%2C+%28sqrt%282.999-x%29%2Fsqrt%282.999-x%29%29%2Aabs%28999%28x-1%29%29+%29+

That triangle has base 1, and the green side is 3, so the vector is 
the hypotenuse and that equals 
 _____    ___    __
Ö1²+3² = Ö1+9 = Ö10

The angle between the vector and the x-axis has a tangent of 3/1, so
the angle is 1.249 radians, approximately.  So the polar form can be 
written any of the following ways:
 __
Ö10[sin(1.249) + j·cos(1.249)]
 __
Ö10 cjs(1.249)
 __
Ö10 Ð 1.249
 
------   

(ii)5 - j2

The picture of 5 - j2 is the vector that 
connects the origin to the point (5,-2) is this vector:

+graph%28+200%2C+200%2C+-2%2C+6%2C+-2%2C+6%2C+-2%2F5%2A%28sqrt%285-x%29%29%2F%28sqrt%285-x%29%29%2Ax%2A%28sqrt%28x%29%29%2F%28sqrt%28x%29%29+%29+

Now we draw in the y-coordinate:

+graph%28+200%2C+200%2C+-2%2C+7%2C+-2%2C+7%2C+-2%2F5%2A%28sqrt%285-x%29%29%2F%28sqrt%285-x%29%29%2Ax%2A%28sqrt%28x%29%29%2F%28sqrt%28x%29%29%2C+-abs%2850%28x-5%29%29+%29+

That triangle has top side 5, and the green side is -2, so the vector 
is the hypotenuse and that equals 
 ________    ____    __
Ö5²+(-2)² = Ö25+4 = Ö29

The angle between the vector and the x-axis has a tangent of -2/5, so
the angle is -.381 radians, approximately.  So the polar form can be 
written any of the following ways:
 __
Ö29[sin(-.381) + j·cos(-.381)]
 __
Ö29 cjs(-.381)
 __
Ö29 Ð -.381

Edwin