trying to solve (j+z)^n + (z-j)^n =0

[pepsi]

hello i'v done the following...

that the moduli of z+j and z-j are the same tells us that z is on the real axis and z+j and z-j are on opposite sides of the real axis.

the figure i have sketched tells me that arg(z+j) = - arg (z-j) so arg(z+j) + arg (z-j) = 0 and arg [(z+j)(z-j)] = arg (x^2-1) =0...
but what good is that??

also b/c (z+j)^n = - (z-j)^n there is a power n such that the sum of z+j raised to n and z-j raised to n is zero... which is whenever the sum of their arguments is (k)pi???

then arg [(z+j)/(z-j)] = arg(z+j) - arg (z-j) = ...?

please help me

 

[Deleted member 4993]

hello i'v done the following...

that the moduli of z+j and z-j are the same tells us that z is on the real axis and z+j and z-j are on opposite sides of the real axis.

the figure i have sketched tells me that arg(z+j) = - arg (z-j) so arg(z+j) + arg (z-j) = 0 and arg [(z+j)(z-j)] = arg (x^2-1) =0...
but what good is that??

also b/c (z+j)^n = - (z-j)^n there is a power n such that the sum of z+j raised to n and z-j raised to n is zero... which is whenever the sum of their arguments is (k)pi???

then arg [(z+j)/(z-j)] = arg(z+j) - arg (z-j) = ...?

please help me

What are you supposed to show - what is the question?

What is z - is it (x + iy)?

What is j?

 

[pepsi]

hi,

and sorry for not being specific, which is completly my fault. j is i (in some circles...) so its just -1^1/2. Also z is indeed z = x +iy

I am actually trying to show that z = cos(k+1)pi/2n for k = 0,1...,n-1 satisfies said equation. This is the book's answer. In the process I would like to understand the question also...