$\because \underset{0}{\overset{\mathrm{x}}{\int }}\log \left|\mathrm{x}\right|\mathrm{dx}=\mathrm{x}(\mathrm{logx}-1)$

$\mathrm{I}=\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\mathrm{sinx}.\log |\mathrm{sinx}-\mathrm{cosx}|\mathrm{dx}$                 .....(1)

$\mathrm{I}=\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\mathrm{cosx}.\log |\mathrm{sinx}-\mathrm{cosx}|\mathrm{dx}$               ......(2)

$\mathrm{I}=\underset{0}{\overset{\mathrm{\pi }/4}{\int }}(\mathrm{sinx}+\mathrm{cosx}).\log |\mathrm{sinx}-\mathrm{cosx}|\mathrm{dx}$

$=\left[(\mathrm{sinx}-\mathrm{cosx})(\log |\mathrm{sinx}-\mathrm{cosx}|-1)\right]{}_0^{\mathrm{\pi }/4}$

Now, Re (x + iy – 1) + Re ((x + iy)²) = –1

            (x – 1) + x² – y² = –1

            x² – y² + x = 0 ⇒ y² = x² + x                               ..............(1)

$\because \left|\mathrm{z}\right|=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}=1\Rightarrow {\text{x}}^{\text{2}}\mathrm{ }+{\text{ y}}^{\text{2}}\mathrm{ }=\text{ 1}$

            x² + x² + x – 1 = 0

            2x² + x – 1 = 0

            (2x –1) (x + 1) = 0         

$\Rightarrow \begin{array}{cc}\mathrm{x}=-1& 1/2\\ \mathrm{y}=0& \sqrt{3}/2\end{array}$

$\therefore \mathrm{z}+\overline{\mathrm{z}}=2\mathrm{x}=1$

$\mathrm{z}+\overline{\mathrm{z}}=2\mathrm{x}=-12$