(1) z=(1-i)^2+1+3i = 1-2i+(-1)+1+3i=1+i|z| = √(1^2+1^2) = √2(2)z^2+az+b = 1-i(1+i)^2 + a(1+i) + b = 1-i2i + a + ai + b = 1-i(a+2)i + (a+b) = 1-i所以 a+b=1,a+2=-1得到 a=-3,b=4