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:
3n + 6 - 8
Step-by-step explanation:
Answer:
its and symobol of an number or letter
Answer:
21
Step-by-step explanation:
Three raise to third power can be written as 
. In this three is the base which is raised to third power.
Next we need to subtract 6 from Three raise to third power(i.e from 27)
which can be written mathematically as
Hence 6 less than three raised to the third power is 21 as explained above.
Answer:
72π (in terms of pi)
Step-by-step explanation:
V = π·r²·h
Plug in the values.
V = π·3²·8
Solve.
V = 72π
To leave an equation in terms of pi, just square, multiply, and leave pi at the end.
3² = 9·8 = 72
72π
If you need the answer fully simplified, it is 226.19 (rounded up).
