\frac{z-i-1}{i(a+ib)+1}=\frac{(a-1)+(b-1)i}{(1-b)+(a+1)i}=\frac{((a-1)+(b-1)i)((1-b)-(a+1)i)}{(1-b)^2+(a+1)^2}=\frac{((a-1)(1-b)+(b-1)(a+1))-(a^2-1+(b-1)^2)i}{(1-b)^2+(a+1)^2}