108=2x-4(2x-9)
108=2x-8x+36
108=-6x+36
-36 -36
72=-6x
72/-6=-6x/-6
-12=x
Given a complex number in the form:
![z= \rho [\cos \theta + i \sin \theta]](https://tex.z-dn.net/?f=z%3D%20%5Crho%20%5B%5Ccos%20%5Ctheta%20%2B%20i%20%5Csin%20%5Ctheta%5D)
The nth-power of this number,

, can be calculated as follows:
- the modulus of

is equal to the nth-power of the modulus of z, while the angle of

is equal to n multiplied the angle of z, so:
![z^n = \rho^n [\cos n\theta + i \sin n\theta ]](https://tex.z-dn.net/?f=z%5En%20%3D%20%5Crho%5En%20%5B%5Ccos%20n%5Ctheta%20%2B%20i%20%5Csin%20n%5Ctheta%20%5D)
In our case, n=3, so

is equal to
![z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]](https://tex.z-dn.net/?f=z%5E3%20%3D%20%5Crho%5E3%20%5B%5Ccos%203%20%5Ctheta%20%2B%20i%20%5Csin%203%20%5Ctheta%20%5D%20%3D%20%285%5E3%29%20%5B%5Ccos%20%283%20%5Ccdot%20330%5E%7B%5Ccirc%7D%29%20%2B%20i%20%5Csin%20%283%20%5Ccdot%20330%5E%7B%5Ccirc%7D%29%20%5D)
(1)
And since

and both sine and cosine are periodic in

, (1) becomes
You can figure the remaining percentage in 2050 and calculate using that, or you can figure the amount of population that will be lost and subtract that from the 2006 value.
<u>First method</u>: The remaining population will be 100% - 5.8% = 94.2% of that in 2006. 0.942×40.2 million ≈ 37.9 million
<u>Second method</u>: 5.8% of 40.2 million is about 2.3 million, so the remaining population will be 40.2 - 2.3 = 37.9 million
Tan ⊕ 9/-9 = -1 so ⊕ = 3π/4
r = √(9^2+(-9)^2) = 12.73
Answer is (12.73, 3π/4)