341Taking modulus on both sides we see that (a^2+b^2)^{2010} = a^2+b^2 so that a^2+b^2 = 0 or a^2+b^2 = 1
So a=b=0 is a solution.
Otherwise let z=a+ib where |z| =1. Then a-ib = 1/z
Hence our equation is z^{2011}=1 which has 2011 distinct solutions.
Hence total number of ordered pairs is 2012