As 2x^4+2+2sqrt3i=02x4+2+2√3i=0
x^4=-1-sqrt3ix4=−1−√3i
Hence x=(-1-sqrt3i)^(1/4)x=(−1−√3i)14
To find this we intend to use DeMoivres Theorem. So let us convert -1-sqrt3i−1−√3i in polar form.
As |-1-sqrt3i|=2∣∣−1−√3i∣∣=2, -1-sqrt3i=2(-1/2-sqrt3/2i)−1−√3i=2(−12−√32i)
or -1-sqrt3i=2(cos(2npi+(5pi)/3)+isin(2npi+(5pi)/3))−1−√3i=2(cos(2nπ+5π3)+isin(2nπ+5π3))
and x=(-1-sqrt3i)^(1/4)=2^(1/4)(cos((2npi)/4+(5pi)/12)+isin((2npi)/4+(5pi)/12))x=(−1−√3i)14=214(cos(2nπ4+5π12)+isin(2nπ4+5π12))
= 2^(1/4)(cos((npi)/2+(5pi)/12)+isin((npi)/2+(5pi)/12))214(cos(nπ2+5π12)+isin(nπ2+5π12))
Now choosing four values of nn i.e. 0,1,20,1,2 and 33, we can get four roots of the equation 2x^4+2+2sqrt3i=02x4+2+2√3i=0, which are
2^(1/4)(cos((5pi)/12)+isin((5pi)/12))214(cos(5π12)+isin(5π12)),
2^(1/4)(cos((11pi)/12)+isin((11pi)/12))214(cos(11π12)+isin(11π12)) i.e. 2^(1/4)(-cos((pi)/12)+isin((pi)/12))214(−cos(π12)+isin(π12))
2^(1/4)(cos((17pi)/12)+isin((17pi)/12))214(cos(17π12)+isin(17π12)) i.e. 2^(1/4)(-cos((5pi)/12)-isin((5pi)/12))214(−cos(5π12)−isin(5π12))
and 2^(1/4)(cos((23pi)/12)+isin((23pi)/12))214(cos(23π12)+isin(23π12)) i.e. 2^(1/4)(cos((pi)/12)-isin((pi)/12))214(cos(π12)−isin(π12))
