DFT basis Function

[panji]

Hi everybody,
I have tried to verify that the DFT basis function
are orthogonal.
If X = (exp(-j*2*pi*n*k/N|k=0:N-1) and Y=(exp(-j*2*pi*m*k/N|k=0:N-1),
Hence, by "Orthogonal" it is mean that
Sigma{k=0:N-1}[(exp(-j*2*pi*n/N) (exp(j*2*pi*n/N)]
= Sigma{k=0:N-1}[(exp(-j*2*pi*(n-m)/N)]
This will be 1 if n=m and 0 if not (n not= m).
But I have failed to verify that
Sigma{k=0:N-1}[(exp(-j*2*pi*n/N) (exp(j*2*pi*n/N)]
= Sigma{k=0:N-1}[(exp(-j*2*pi*(n-m)/N)]=0, for n not=m
Is there anyway to see basis function of DFT is orthogonal especially for n not= m?

[Baraa Alyousef]

HI Mr.
as U know the result of 2*x (whatever x was) is even, then if we apply that on the situation of n not= m we will have an even number in the exp(j*even*pi)
we know that exp(x) = cos(x) + j sin(x) if x=even*pi then sin () = 0 and cos = 1 ..... simple subistitution in DFT formula U will get the answer of Ur question ....

sincerely.
Bara'a Alyousef