Find the inverse of the function y = cube root (x/3)-1
1 answer:
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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)
The nth-power of this number,
- the modulus of
In our case, n=3, so
And since
and both sine and cosine are periodic in