Q13/041)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 ans