Evaluate

This isnt that tough.

Try it like this...

1+i = √2 ( 1/√2 + i/√2 ) = √2 (cos pi/4 + i sin pi/4) = √2 e^{i Π/4}

1-i = √2 ( 1/√2 - i/√2 ) = √2 (cos -pi/4 + i sin -pi/4) = √2 e^{-i Π/4}

Now i think it is simple to solve this question... if u cant.. tell me :)

now take 2 cases n even & n odd

for n even , it will be

1-i =2^n/2 ( e^{n i Î /2} + e^{-i Î /2})

so cos (npi/2)+i sin (npi/2) = cos (npi/2)

same way the other case for odd n..

yes it is valid :)