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
![z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]](https://tex.z-dn.net/?f=z%5E3%20%3D%20125%20%5B%5Ccos%20270%5E%7B%5Ccirc%7D%20%2B%20i%20%5Csin%20270%5E%7B%5Ccirc%7D%20%5D)
Answer:
12.56 yards
Area of Triangle: 
So, 
Finally, 3.14*4 = 12.56
Step-by-step explanation:
Answer: The second one
Step-by-step explanation:
9514 1404 393
Answer:
23) x = ±3i, ±√2
26) x = 4/3, (-2/3)(1 ± i√3)
Step-by-step explanation:
23) Put in standard form to make factoring easier.
x^4 +7x^2 -18 = 0
(x^2 +9)(x^2 -2) = 0 . . . . factors in integers
Using the factoring of the difference of squares, you can continue to get linear factors in complex and irrational numbers:
(x -3i)(x +3i)(x -√2)(x +√2) = 0
x = ±3i, ±√2
___
26) This will be the difference of cubes after you remove the common factor.
81x^3 -192 = 0
3(27x^3 -64) = 0
(3x -4)(9x^2 +12x +16) = 0 . . . . . factor the difference of cubes
The complex roots of the quadratic can be found using the quadratic formula.
x = (-12 ±√(12^2 -4(9)(16)))/(2(9)) = (-12 ±√-432)/18 = -2/3 ± √(-4/3)
Then the three solutions to the equation are ...
x = 4/3, (-2/3)(1 ± i√3)
