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Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

how do i convert the equations like rectangular form on the left to polar form. What are the polar form equations for the 5 left equations?

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  • Alan
    Lv 7
    2 months ago

    (3 of 5 so far) can't do the 5th 

    Summary   

    1st  a. 5.1<348

    2nd.  b.  1.09 < 350 

    3rd. c.  1037 < 15 

    4th.  d.  1.51 < 170 (  also answers 1.51<50, 1.51< 290)  

    1st one (checking ) 

    j^2 = -1 

    j^3 =  -j

    j^4 =  1 

    j^5  = j  

    j^25 = j^5 *j^5*j^5 *j^5 *j^5  = j^5  = j 

    j^20 = j^5*j^5*j^5*j^5  = j^4  = 1

    j^15 = j^3 =  -j

    j^10 = j^2 =    -1 

    therefore 

    j  + 3 +  -2j + 1  + 1 =  -j + 5 

    magnitude   =   sqrt(26) =  5.099019514

    = atan2( 5, -1)  or atan( -1/5) in q4 

    gives  -11.30993247

    or approx.348.7  degrees

    5.099019514 < 348.7 degrees , the 1st answer on the 

    right 

    A.   5.1<348 is the closest

    2nd one 

    2nd term

    j/e^(pi/2)*j = j/j = 1

    so 2nd term = 1

    e^j = cos(1) + jsin(1)

    use conjugate to clear imaginary value from 1st term

    (cos(1) + jsin(1)) (-3-4j)/ ((-3 + 4j) (-3 -4j) )

    denominator

    9 + (4j)(-4j) = 9 - 16j^2 = 25

    (cos(1)+ j*sin(1) ) (-3 -4j)/ 25

    (cos(1) + j(sin(j)) (-3- 4j) /25

    combining terms

    (cos(1) + jsin(1)) (-3-4j) / 25 + 25/25

    (-3*cos(1) - 3jsin(1) - 4j*cos(1) -4*(j^2)sin(1)+ 25) / 25

    (-3*cos(1) + 4*sin(1) + 25 - j*(3sin(1) - 4cos(1)) ) /25

    ( (-3/25)cos(1) +(4/25)sin(1) + 1 + j ((3/25)sin(1) + (4/25)cos(1))

    in decimal multiplied out

    1.075868566+ -0.203722703j

    magnitude = sqrt( 1.075868566^2 + 0.203722703^2)

    magnitude = 1.094986809

    angle = atan2( 1.075868566, 0.203722703) =349.2776225 degrees 

    b. 1.09 < 350  is the closest 

    3rd one

    ( cos(1/2) + jsin(1/2) -4 - 3j ) ^(5) 

    (  (-4 + cos(1/2)+ j ( sin(1/2) -3) )^5 

    ( -3.122417438 +  -2.520574461 j)^5 

    ( sqrt( -3.122417438^2  + -2.520574461^2) 

    <atan2(-3.1224, -2.520))^5 

    (  4.012827715 < -141.0877636 )^5  

    ( 4.012827715 <218.9122364 )^5 

    4.012827715^5 <  (218.9*5)

    1040.525126 < 1094.561182

    1040.525126 < (1094.561182-3*360) 

    1040.525126 < (14.561 )

    which rounds to 

    1041 < 15  

    which is closest to 

    c. 1037 < 15 

    4th one 

    (  -4  +  2*cos(60) + j*2*sin(60) )^(1/3) 

    (-4  +1  + j*2*sqrt(3)/2) )^(1/3) 

    (-3 + j*sqrt(3))^(1/3) 

    ( sqrt( 9 +  3) < atan2(-3, sqrt(3)))^(1/3) 

    (sqrt(12))^(1/3)  < (atan(-3,sqrt(3))/3

    1.513085749 <  (150+360k)/3) = 

    1.513085749 < (50 + 120k)  

    so 

    1.51 < 50 

    1.51 < 170

    1.51 <  290 

    1.51 <  50 

    1.51 < 170 

    1.51 < 290 

    and so on 

    D. 1.51<170 is the closest 

    • Alan
      Lv 7
      2 months agoReport

      I meant (4 of 5 so far) can't do the 5th

      Surprising the answers are a,b,c,d 
      then, if you're lucky the
      5th will be e 
      Not enough information to do the 5th 
      You need Z_3 when the question uses Z_3 
      We know nothing about Z_3 

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  • 2 months ago

    To convert polar to rectangular                 

    x = r cos(theta)                

    y = r sin(theta)                

                    

    so     r    theta    x    y

    A    5.1    349    5.01    -0.97

    B    1.09    350    1.07    -0.19

    C    1037    15    1001.67    268.40

    D    1.51    170    -1.49    0.26

    E    1.41    285    0.36    -1.36

    F    1037    219    -805.90    -652.61

    G    1.41    105    -0.36    1.36

    H    5.1    191    -5.01    -0.97

                    

                    

    j is used by engineers to represent the square root of minus 1.  ("i" is used for current.)                

    j^25 = j^24*j =( j^4)^6 * j = 1^6 * j = 1*j = j                

    3 j^20 = 3* (j^4)^5 = 3 * 1 ^5 = 3                

    2 j^15 = 2 *( j^4)^3 * j^3 = 2* j^3 = -2j                

    -j^10 = - (j^4)^2 * j*2 = - (-1) = 1                

    1 = 1                

    the sum is                

    5 - j                

                    

    We cannot find Z-sub-T because Z-sub-3 is not defined.                

                    

    e^(j*theta) = cos(theta) + j sin(theta)   theta in radians                

    so                 

    e^j = cos(1) + j sin(1)                

                    

    That is 9 problems and a bit more for 2 points.                

                    

    I hope this helps.                

    • ...Show all comments
    • what are the answers of the equations on the left to polar form. When you convert the equations on the left, it should equal one of the polar form equations on the right. I am trying to check my answers

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