![z^{1/n} = r^{1/n}\cdot \left[\cos \left(\frac{\theta + 2\cdot k\cdot \pi}{n} \right)+i\,\sin \left(\frac{\theta + 2\cdot k\cdot \pi}{n}\right)\right]](/tpl/images/1154/1715/c5478.png)
, for 
(1)
- Grade of the root. 
- Norm of the complex number.
- Direction of the complex number, measured in radians.
- Index of the solution. 
, then the root of the number is:
, then the root of this number is:![z^{1/3} = 2^{1/6}\cdot \left[\cos \left(\frac{\frac{3\pi}{4} }{3} \right)+i\,\sin \left(\frac{\frac{3\pi}{4} }{3} \right)\right]](/tpl/images/1154/1715/6dcf3.png)
![z^{1/3} = 2^{1/6}\cdot \left[\cos \frac{\pi}{4}+i\,\sin \frac{\pi}{4} \right]](/tpl/images/1154/1715/aeaf9.png)
![z^{1/3} = 2^{1/6}\cdot \left[\cos 45^{\circ}+i\,\sin 45^{\circ}\right]](/tpl/images/1154/1715/9552b.png)
Find the root for (-1+i)^1/3
...
Mathematics, 27.02.2021 14:00 ShadowShu8658
Find the root for (-1+i)^1/3
Answers: 2
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