Question

If n is positive number then prove that
(1+i)^n+(1-i)^n=2^n/2+1 cosnpi/4

Answer 1

Answer:

| 1 + i | = √( 1² + 1² ) = √2

arg ( 1 + i ) = π/4     ( it's a 45° angle from the origin to 1+i )

=> 1 + i = √2 exp ( iπ/4 )

Similarly (or by conjugation)

1 - i = √2 exp ( -iπ / 4 )

So...

( 1 + i )ⁿ + ( 1 - i )ⁿ

= [ √2 exp ( iπ/4 ) ]ⁿ  +  [ √2 exp ( -iπ/4 ) ]ⁿ

= 2^(n/2) exp ( inπ/4 )  +  2^(n/2) exp ( -inπ/4 )

= 2^(n/2) [ cos nπ/4 + i sin nπ/4 ] + 2^(n/2) [ cos nπ/4 - i sin nπ/4 ]

= 2^(n/2) [ 2 cos nπ/4 ]

= 2^(n/2 + 1) cos nπ/4