Let $Z\in C$  (set of complex numbers)
A and B be two sets are defined as
$\text{A }=\left\{\text{ Z }/\left|\text{Z }–\text{ 1}\right|=\text{ 2}\sqrt{2}\right\}$
$B=\left\{Z/\frac{Z-1}{2}-\frac{2\sqrt{2}}{\overline{Z}-1}=\frac{\lambda e^{-i\pi /4}}{|Z-1|},where\lambda isrealparameter\right\}$
Then the sum of real parts of all complex numbers in $A\cap B$ is 

$|Z-1|=2\sqrt{2}$
 $$\begin{gathered} \text{ Let }\mathrm{Z}-1=2\sqrt{2}{\mathrm{e}}^{\mathbf{i}\mathit{\theta }} \\ \Rightarrow \frac{Z-1}{2}=\sqrt{2}{e}^{i\theta }\Rightarrow \frac{\overline{Z}-1}{2}=\sqrt{2}{e}^{-i\theta }\to \left(1\right) \\ \therefore \frac{2\sqrt{2}}{\overline{Z}-1}=e^{i\theta }\to \left(2\right) \end{gathered}$$    
Substitute(1), (2) in (B), we get
 $$\begin{gathered} \sqrt{2}{e}^{i\theta }-e^{i\theta }=\frac{\lambda e^{-i\pi /4}}{2\sqrt{2}}\Rightarrow e^{i\theta }(\sqrt{2}-1)=\frac{\lambda e^{-i\pi /4}}{2\sqrt{2}} \\ \Rightarrow {e^i}^{(\theta +\pi /4)}=\frac{\lambda }{2\sqrt{2}(\sqrt{2}-1)}=\mathrm{Re}al \\ \Rightarrow Sin\left(\theta +\frac{\pi }{4}\right)=0\Rightarrow \left(\theta +\frac{\pi }{4}\right)=0or\pi \\ \theta =\frac{-\pi }{4}or\frac{3\pi }{4} \end{gathered}$$

 Z = 3 – 2i   or   –1 + 2i
Sum of real parts 2