There is no general solution for `ninNN.`
For `n=0:`
`(x+i)^0-(x-i)^0=0`
`1-1=0`
Which is valid for all real numbers hence `x inRR.`
For `n=1:`
`(x+i)^1-(x-i)^1=0`
`x+i-x+i=0`
`2i=0`
Which is never true hence there is no solution.
For `n=2:`
`(x+i)^2-(x-i)^2=0`
`x^2+2x i-1-(x^2-2x i-1)=0`
`4x i=0`
`x=0`
Obviously I could go on like this...
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
There is no general solution for `ninNN.`
For `n=0:`
`(x+i)^0-(x-i)^0=0`
`1-1=0`
Which is valid for all real numbers hence `x inRR.`
For `n=1:`
`(x+i)^1-(x-i)^1=0`
`x+i-x+i=0`
`2i=0`
Which is never true hence there is no solution.
For `n=2:`
`(x+i)^2-(x-i)^2=0`
`x^2+2x i-1-(x^2-2x i-1)=0`
`4x i=0`
`x=0`
Obviously I could go on like this forever. E.g. for `n=3` solutions are `x_(1,2)=pm1/sqrt3.` But the point is that there are no general solution when `n` is an integer (or even worse a real number). Another way to prove this is to use de Moivre's formula.