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
Glad you asked this question! Over the years of working with my professor we learned that if phs= 5 (constant) times the j to the power of 2 you will get the formula or polynomialic equation. It quite fascinating and due to using his equation the answer in fact is "c". Cheers
im assuming i just pick two angles cause i cant see the options? two 90 degree angles.
and my name is teagan too :D
500 − 250 = 250 pieces left after donation
Yolanda got 180 pieces....then Enrique got 70 pieces
So he got 70 /250 = 7/25 = 28% of the remaining candy
Answer:
I think it's 4
Step-by-step explanation:
Three resistors with values of 12 Ω, 24 Ω and 6 Ω are connected in series to one another and a 24 V power supply. Draw the circuit and fill in the data table.