1.   determine the cubic roots of 8(cos120 + jsin120) 2.   solve for x & y if y(j2-2) - x(j4 + 3) = 2In problem 1 write down the principal root in the form a + bj with a & b real numbers.

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kjcdb8er | Teacher | (Level 1) Associate Educator

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1. It's easiest to find the cubic root of 8(cos120 + jsin120) if we convert it into polar form (via euler's rule). We need to use radians, not degrees for this: 120o = 2pi/3 rad

8(cos120 + jsin120) = 8(cos 3pi/2 + jsin 3pi/2) = 8 exp(j3pi/2)

Now apply the cubic root:

(8 exp(j3pi/2))^1/3 = 8^1/3 exp(j3pi/2 * 1/3)

2exp(jpi/2)

to convert the above into rectangular form, use euler's identity:

2exp(jpi/2) = 2(cos pi/2 + jsin pi/2) = 2j

2. If x and y are not real numbers, then you don't have enough information to solve this problem. So, lets assume that x and y are real.

y(2j-2) - x(4j+3) = 2

2jy - 2y - 4jx - 3x = 2 + 0j

This system gives you two equations, since the imaginary and real components have to sum to 0 and 2, respectively.

2jy - 4jx = 0     and      -2y - 3x = 2

x = y/2           ---->      -2y - 3(y/2) = 2

.                                   -y(7/2) = 2

y = -4/7 and x = -2/7

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