Express √3 + i in polar form:
|√3 + i| = √((√3)² + 1²) = √4 = 2
arg(√3 + i) = arctan(1/√3) = π/6
Then
√3 + i = 2 (cos(π/6) + i sin(π/6))
By DeMoivre's theorem,
(√3 + i)³ = (2 (cos(π/6) + i sin(π/6)))³
… = 2³ (cos(3 • π/6) + i sin(3 • π/6))
… = 8 (cos(π/2) + i sin(π/2))
… = 8i
Since one equation has a negative y and the other has a positive y, I'm going to use those since they cancel each other out. Before that, the two y's need to be equal to each other.
x+2y=6
x-y=3
Multiply the bottom equation by two so then you have:
x+2y=6
2x-2y=6
The y's now cancel out:
x=6
2x=6
Add them together
3x=12
Divide
x=4.
To find y, plug x into either equation (*don't have to do both, but I will)
(4)+2y=6
(4)-y=3
Subtract four
2y=2
-y=-1
Divide each
2y/2 = 2/2
y=1
-y/-1 = -1/-1
y=1
The answer is:
x=4
y=1
I hope that helps!
D. -15
All I did was plug the numbers in
8 x 64
=64 x 2 x 4
=128 x 2 x 2
=256 x 2
=512
