2011/010) Find integral solution of (1-i)^n = 2^n

We have 1-i = (sqrt(2) cis (-pi/4)

So (1-i)^n = 2^(n/2) cis (-npi/4) = 2^n

The modulo of LHS = 2^(n/2) and rhs = 2^n and both are same

If 2^(n/2) = 2^n or n = 0

Then = 2^(n/2) cis (-npi/4) = 1 = RHS

So n = 0 is the only solution