Show that for all n N.
To show:
Taking LHS,
[rationalize]
[∵ i2 = -1]
= (1 – i)n(1 + i)n
= [(1 – i)(1 + i)]n
= [(1)2 – (i)2]n [(a + b)(a – b) = a2 – b2]
= (1 – i2)n
= [1 – (-1)]n[∵ i2 = -1]
= (2)n
= 2n
= RHS
Hence Proved