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...

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