# 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

Asked on by helga95

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll re-write the equation:

`(x+i)^n + (x-i)^n = 0`

We'll divide by `(x-i)^n`  both sides:

`((x+i)/(x-i))^n + 1 = 0`

We'll subtract 1 both sides:

`((x+i)/(x-i))^n = -1` We'll write -1 as a trigonometric form of a complex number:

`-1 = cos pi + isin pi`

We'll re-write the equation:

`((x+i)/(x-i))^n = cos pi + isin pi`

We'll remove the n-th power from the left side, raising both sides to the power `(1/n):`

`` `(x+i)/(x-i) = (cos pi + isin pi)^(1/n)`

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

`(x+i)/(x-i) = (cos ((2k+1)pi/n) + isin ((2k+1)pi/n))` k is an integer number.

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

`(x+i)^2/((x+i)(x-i)) = (cos ((2k+1)pi/n) + isin ((2k+1)pi/n))`

The difference of squares from denominator returns the product `x^2 + 1` .

We'll cross multiply and we'll get:

`x^2 + 2ix - 1 = (x^2 + 1)(cos ((2k+1)pi/n) + isin ((2k+1)pi/n))`

`x^2 + 2ix - 1 = (x^2 + cos ((2k+1)pi/n)) + i(x^2*sin ((2k+1)pi/n) + sin ((2k+1)pi/n))`

We'll compare both sides:

`x^2 - 1 = x^2 + cos ((2k+1)pi/n)`

`` `cos ((2k+1)pi/n) = -1`

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

`sin[((2k+1)pi/n]= (2x)/(x^2 + 1)`

`x_k = cot [((2k+1)pi)/(2n)]`

Therefore, the solutions of the given equation are: `x_k = cot [((2k+1)pi)/(2n)].`

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