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? Relevance
• (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

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

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

• Login to reply the answers
• 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.

• Login to reply the answers