Let OA be the Line from Z=0 to Z= 1+i

1) On the line OA i.e y=x

diff.

dy=dx---(1)

z=x+iy

diff.

dz =dx+ idy

dz=dx+idx----from equ.(1)

dz = (1+i) dx-----(2)

x varies from 0 to 1..

$I=\int_{0}^{1+i}(z)^2$

$\int_{0}^{1+i} (x+iy)^2 dz$

$\int_{0}^{1}(x^2+2xyi+i^2y^2)(1+i)dx$----from eq.(2)

$\int_0^1 (x^2+2xiy-y^2)(1+i)dx$

since y = x

$\int_0^1 (x^2+2xix-x^2)(1+i)dx$

$\int_0^1(i2x^2)(1+i)dx$

$(1+i)2i \int_0^1 x^2 dx$

$(1+i)2i [\frac {3}{x^3}]_0^1$

$(1+i)2i [\frac {1}{3}-0]_0^1$

$I=\frac{2}{3}(i-1)$

2) On the arc OA on the parabola

$x=y^2$------(3)

diff.

dx=2y dy-----(4)

z = x+iy

diff.

dz =dx + idy

dz = 2ydy+idy -----------from equation (4)

dz =(2y+i)dy----(5)

$I=\int_0^1(z^2)dz$

=$\int_0^1(x+iy)^2(2y+i)dy$

$=\int_0^1(x^2+2ixy-y^2)^(2y+i)dy$

from eqn(3), $x=y^2$

$=\int_0^1(y^4+2y^2 iy-y^2)(2y+i)dy$

$=\int_0^1(y^4+2y^3 i-y^2)(2y+i)dy$

$=\int_0^1(2y^5+4y^4 i-2y^3+iy^4 - 2y^3 -iy^2)dy$

$=\int_0^1(2y^5+5y^4 i-4y^3-iy^2)dy$

$=\int_0^1[(2y^5-4y^3) +i(5y^4-y^2)]dy$

$=[(\frac{2y^6}{6}-\frac{4y^4}{4})+ i(\frac{5y^5}{5}-\frac{y^3}{3})]_0^{1}$

upper limit - lower limit...

$=[\frac{2}{6}-\frac{4}{4}+i(\frac{5}{5}-\frac{1}{3})]-[(0-0]+i(0-0)]$

$=[\frac{1}{3}-1]+i[1-\frac{1}{3}]$

$I = \frac {2}{3}[i-1]$

The two integral are equal.

i.e integral is independent of path because

f(z) =$z^2$ is an analytic function.