If

, then


and

.
Then


.
Since
![\theta \in (-\pi,\pi]](https://tex.z-dn.net/?f=%5Ctheta%20%5Cin%20%28-%5Cpi%2C%5Cpi%5D)
, you can conclude that

.
The triginometric form of z is

.
Use
formula
![\sqrt[3]{z} =\{ \sqrt[3]{|z|} (cos \frac{\theta+2\pi k}{n}+isin \frac{\theta+2\pi k}{n} ), k=0,1,2\}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bz%7D%20%3D%5C%7B%20%5Csqrt%5B3%5D%7B%7Cz%7C%7D%20%28cos%20%5Cfrac%7B%5Ctheta%2B2%5Cpi%20%0Ak%7D%7Bn%7D%2Bisin%20%5Cfrac%7B%5Ctheta%2B2%5Cpi%20k%7D%7Bn%7D%20%29%2C%20k%3D0%2C1%2C2%5C%7D)
to find cube
root.
For k=0,
![z_1= \sqrt[3]{2} (cos \frac{ \frac{2\pi}{3} }{3} +isin\frac{ \frac{2\pi}{3} }{3})= \sqrt[3]{2} (cos 40^0+isin 40^0)](https://tex.z-dn.net/?f=z_1%3D%20%5Csqrt%5B3%5D%7B2%7D%20%28cos%20%5Cfrac%7B%20%0A%5Cfrac%7B2%5Cpi%7D%7B3%7D%20%7D%7B3%7D%20%2Bisin%5Cfrac%7B%20%5Cfrac%7B2%5Cpi%7D%7B3%7D%20%7D%7B3%7D%29%3D%20%5Csqrt%5B3%5D%7B2%7D%20%28cos%20%0A40%5E0%2Bisin%2040%5E0%29)
,
For k=1,
![z_1= \sqrt[3]{2} (cos \frac{ \frac{2\pi}{3}+2\pi }{3} +isin\frac{ \frac{2\pi}{3}+2\pi }{3})= \sqrt[3]{2} (cos 160^0+isin 160^0)](https://tex.z-dn.net/?f=z_1%3D%20%5Csqrt%5B3%5D%7B2%7D%20%28cos%0A%20%5Cfrac%7B%20%5Cfrac%7B2%5Cpi%7D%7B3%7D%2B2%5Cpi%20%7D%7B3%7D%20%2Bisin%5Cfrac%7B%20%5Cfrac%7B2%5Cpi%7D%7B3%7D%2B2%5Cpi%20%7D%7B3%7D%29%3D%20%0A%5Csqrt%5B3%5D%7B2%7D%20%28cos%20160%5E0%2Bisin%20160%5E0%29)
,
For k=2,
![z_1= \sqrt[3]{2} (cos \frac{ \frac{2\pi}{3}+4\pi }{3} +isin\frac{ \frac{2\pi}{3}+4\pi }{3})= \sqrt[3]{2} (cos 280^0+isin 280^0)](https://tex.z-dn.net/?f=z_1%3D%20%5Csqrt%5B3%5D%7B2%7D%20%28cos%20%5Cfrac%7B%20%5Cfrac%7B2%5Cpi%7D%7B3%7D%2B4%5Cpi%20%7D%7B3%7D%20%2Bisin%5Cfrac%7B%20%0A%5Cfrac%7B2%5Cpi%7D%7B3%7D%2B4%5Cpi%20%7D%7B3%7D%29%3D%20%5Csqrt%5B3%5D%7B2%7D%20%28cos%20280%5E0%2Bisin%20280%5E0%29)
.
Answer: The correct choice is C.
Answer:
(-2,-3)
Step-by-step explanation:
look at the pic for a little bit of an understanding.
i hope this helps
have a nice day/night
mark brainliest, please :)
Answer:
b
Step-by-step explanation:
4.003 4.03 4.033 4.3 4.303 4.33