I need to solve the equation ((x+i) raise to n)+((x-i) raise to n)=0. i must use trigonometry but i don't know how. thanks

We'll re-write the
equation:

0

We'll divide by   both
sides:

0

We'll subtract 1 both
sides:

We'll write -1 as a
trigonometric form of a complex number:

pi

We'll re-write the
equation:

pi

We'll remove the n-th power from the left side, raising
both sides to the power 

 
+ isin pi)^(1/n)

We'll apply Moivre's rule to the right
side:

((2k+1)pi/n))

We'll multiply the
denominator from the left side, by it's
conjugate:

((2k+1)pi/n))

The difference of squares from denominator
returns the product .

We'll cross multiply and
we'll get:

isin ((2k+1)pi/n))

((2k+1)pi/n)) + i(x^2*sin ((2k+1)pi/n) + sin
((2k+1)pi/n))

We'll compare both
sides:

((2k+1)pi/n)

 
-1

 
((2k+1)pi/n)

1)

[((2k+1)pi)/(2n)]

Therefore, the solutions
of the given equation are: