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tatyana61 [14]
3 years ago
10

Which expression is a cube root of -1+i sqrt3

Mathematics
1 answer:
Marina CMI [18]3 years ago
4 0
If z=-1+i \sqrt{3}, then 

Rez=-1
Imz= \sqrt{3} and 
|z|= \sqrt{Re^2z+Im^2z}= \sqrt{1^2+ (\sqrt{3})^2 }= \sqrt{4}=2.
Then cos\theta = \frac{Rez}{|z|} = \frac{-1}{2}sin\theta = \frac{Imz}{|z|}= \frac{ \sqrt{3} }{2}.

Since \theta \in (-\pi,\pi], you can conclude that \theta= \frac{2\pi}{3}.
The triginometric form of z is z=|z|(cos\theta+isin\theta)=2(cos \frac{2\pi}{3} +isin \frac{2\pi}{3} ).
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\} 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),
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),
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).
Answer: The correct choice is C.
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