Question

How do you solve `2x^4+2+2sqrt3i=0`?

Answer 1

`x=2^(1/4)(cos((5pi)/12)+isin((5pi)/12))` or `2^(1/4)(-cos((pi)/12)+isin((pi)/12))` or `2^(1/4)(-cos((5pi)/12)-isin((5pi)/12))` or `2^(1/4)(cos((pi)/12)-isin((pi)/12))`

Explanation:

As `2x^4+2+2sqrt3i=0`

`x^4=-1-sqrt3i`

Hence `x=(-1-sqrt3i)^(1/4)`

To find this we intend to use DeMoivres Theorem. So let us convert `-1-sqrt3i` in polar form.

As `|-1-sqrt3i|=2`, `-1-sqrt3i=2(-1/2-sqrt3/2i)`

or `-1-sqrt3i=2(cos(2npi+(5pi)/3)+isin(2npi+(5pi)/3))`

and `x=(-1-sqrt3i)^(1/4)=2^(1/4)(cos((2npi)/4+(5pi)/12)+isin((2npi)/4+(5pi)/12))`

= `2^(1/4)(cos((npi)/2+(5pi)/12)+isin((npi)/2+(5pi)/12))`

Now choosing four values of `n` i.e. `0,1,2` and `3`, we can get four roots of the equation `2x^4+2+2sqrt3i=0`, which are

`2^(1/4)(cos((5pi)/12)+isin((5pi)/12))`,

`2^(1/4)(cos((11pi)/12)+isin((11pi)/12))` i.e. `2^(1/4)(-cos((pi)/12)+isin((pi)/12))`

`2^(1/4)(cos((17pi)/12)+isin((17pi)/12))` i.e. `2^(1/4)(-cos((5pi)/12)-isin((5pi)/12))`

and `2^(1/4)(cos((23pi)/12)+isin((23pi)/12))` i.e. `2^(1/4)(cos((pi)/12)-isin((pi)/12))`