(1)z=t+t^2i dz=(1+2ti)dt ∫zdz =∫[0→1](t+t^2i)(1+2ti)dt =∫[0→1](t-2t^3+t^2i+2t^2i)dt =∫[0→1](t-2t^3)dt+i∫[0→1]3t^2dt =[(1/2)・t^2-(1/2)t^4][0→1]+i[t^3][0→1] =i・・・答え (2) z=itより dz=idt (1+z)^2=(1+it)^2 ∫(1+z)^2dz =∫[0→1]i(1+it)^2dt =∫[0→1]i(1+2it-t^2)dt =∫[0→1]-2tdt+i∫[0→1](1-t^2)dt =[-t^2][0→1]+i[t-(1/3)・t^3][0→1] =-1+(2/3)i・・・答え