$\int_0^{2+i}{(\overline{z})^2}dz $ along

i. Y = x/2

ii. The real axis to 2 and then vertically to 2 + i

let $z = x+iy - - - - 1$

given $ y = x/2 - - - - 2$

diff 1 w.r.t x

$dz = dx + idy $

but from equation 2

$ \begin{equation} y = x/2\\ \therefore dy = dx/2\\ \therefore dz = dx + i\ dx/2\\ dz = (1 + i/2)dx - - - - 3\\ \end{equation} $

1. For first part

$\begin{equation} \overline {z} = x- iy, \\ \int_0^{2+i} {(\overline {z})^2}dz = \int_0^{2+i}{(x-iy)^2(1+i/2) dx} \\ \text{Here 'x' varies from 0 to 2} \\ \therefore = \int_0^{2}{(x-iy)^2(1+i/2) dx} \\ \therefore = \int_0^{2}{(x^2-2ixy-y^2)(1+i/2) dx} \\ \therefore = \int_0^{2}{(x^2-2ixy-y^2 + ix^2/2 + 2xy/2 -iy^2/2) dx} \\ \therefore = \int_0^{2}{(x^2 +xy-y^2 + i (x^2/2 + 2xy -y^2/2) dx} \\ \text{Now, put y= x/2}\\ \therefore = \int_0^{2}\{{(x^2 - x^2/4 + x^2/2 + i (x^2/2 - 2x^2/2 -x^2/8) dx} \}\\ \therefore = \int_0^{2}{[5/4 x^2 + i(-5/8 x^2)]dx} \\ \therefore = (5/4-5i/8 \int_0^{2}{x^2 dx} \\ \therefore = (5/4-5i/8) [x^3/3]_0^2 \\ \text{upper limit - lower limit} \\ \therefore = (5/4 - 5i/8 )(8/3 - 0)\\ = 10/3 - 5i/3 \end{equation}$

ii. $\begin{equation} z= x+iy\\ \text{First we find real axis o to 2,}\\ \therefore y = 0 \\ z = x+iy \\ z= x \\ \frac{dz}{dx} = dx \\ \int_0^{2+i}{(\overline{z}^2)}dz = \int_0^2{(x-iy)^2}dx\\ \int_0^{2+i}{(x)^2} = [\frac{x^3}{3}]_0^2 = \frac83\\ \text{Now find along vertical axis, 0 to (2+i). Here vertical axis varies from 0 to 1}\\ \therefore z = x+iy, \ \ \ \& \ \overline{z} = x-iy \\ dz = idy\\ \int_0^{2+i}{(\overline{z}^2)}dz = \int_0^2{(x-iy)^2}idy\\ \int_0^1{(x^2-2ixy-y^2)}idy= \int_0^1{(2^2-2i2y-y^2)}idy \\ = \int_0^1{(4-4iy-y^2)}idy \\ = \int_0^1{(4i-4y-y^2)}idy \\ =[4iy+4y^2/2-iy^3/3]_0^1 \\ = [(4i+4/2-i/3)-(0+0_0)] = (2+11i/3) \\ \text{Hence adding the two result} \\ I = 8/3+2+11i/3 \\ I = 14/3+11i/3 \\ \end{equation}$