Answers: 1
, then 
and 
.
.![\theta \in (-\pi,\pi]](/tpl/images/0101/2196/a3daf.png)
, you can conclude that 
.
.![\sqrt[3]{z} =\{ \sqrt[3]{|z|} (cos \frac{\theta+2\pi k}{n}+isin \frac{\theta+2\pi k}{n} ), k=0,1,2\}](/tpl/images/0101/2196/4dd91.png)
to find cube
root. ![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)](/tpl/images/0101/2196/1b1b6.png)
,![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)](/tpl/images/0101/2196/4bb5c.png)
,![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)](/tpl/images/0101/2196/f18d6.png)
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