Is (-i)^(-m) Equal to cos((m*pi)/2)+i*sin((m*pi)/2) in Complex Analysis?

[dado033]

is this relashion true? or false?
if it is true how can I proof it?
(-i)^(-m) = cos((m*pi)/2)+i*sin((m*pi)/2)

 

[John Creighto]

Perhaps try showing the left hand side is periodic. Then I think this would mean that you would only have to work out a finite number of cases.

 

[mathman]

is this relashion true? or false?
if it is true how can I proof it?
(-i)^(-m) = cos((m*pi)/2)+i*sin((m*pi)/2)

-i=exp(-i(pi/2))
Therefore (-i)^(-m)=exp(i(m*pi/2))
Now apply Euler's identity.

 

[dado033]

thank u very much