If the curve is not a closed curve then to evaluate the value of integral

we use Cartesian form of z i.e. z = x + iy i.e. dz = dx + idy ,

the path of the integration is the line joining points z=0 i.e. (0,0) to z= 1+I i.e. (1 ,1) given as

$ \frac{(y-0)}{(1-0)}=\frac{(x-0)}{(1-0)}$ or y = x

Putting above values in the given integral,

value of the integral along the line y=x i.e. dy =dx is given by:

$I = ∫_0^1(x-x+ix^2) (dx+ idx)$

$= (1+i) ∫_0^1ix^2 dx = (i -1) [\frac{x^3}{3}]_0^1 = \frac{(i -1)}{3}$