Hi Chunhak, thank you for your question. I would choose the new variable w=z-1-i; then you need to show that f(w)=(w+1+i)^2=w^2+2w(1+i)+2i approaches 2i as w goes to zero. And that is easier (use triangle inequality for |f(w)-2i| and pick delta small enough s.t. epsilon>sqrt(2)delta+delta^2, then you have |f(w)-2i|
