We've used $z=i(Z+4/Z)$ as a conformal mapping to map the exterior of a circle $|Z|=2$ to the exterior of the line segment $(-4i,4i)$.
We now want to write the complex potential of the uniform flow past the flat plate, $x=0, -4<y<4$, parallel to it
Any help is appreciated, we're all pretty stuck here, so thanks
So for potential flow $\nabla^2 \Phi = 0$ where $\vec v = \nabla \Phi$. 2D potential flow problems can be solved with complex analytic functions because if $F(x + i y) = \phi(x, y) + i \psi(x, y)$ is analytic (for functions $\phi,\psi: \mathbb R^2 \rightarrow \mathbb R$) then the Cauchy-Riemann conditions give us $\nabla^2\phi = 0 = \nabla^2 \psi$. Furthermore, to get the real "meaning" of this, consider how $F$ responds to a change $dz = dx + i ~dy$: $$dF = (\partial_x \phi) ~dx + (\partial_y \phi) ~ dy + i ~ (\partial_x \psi) ~ dx + i ~(\partial_y \psi)~ dy.$$Replacing $\psi$ with $\phi$ via the Cauchy-Riemann conditions gives$$dF = (\partial_x \phi) ~dx + (\partial_y \phi) ~ dy + i ~ (-\partial_y \phi) ~ dx + i ~(\partial_x \phi)~ dy$$$$\frac{dF}{dz} = \partial_x \phi - i \partial_y \phi $$So $\phi(x, y) = \Phi(x, -y)$ gives $\frac{dF}{dz} = v_x + i ~ v_y$.
For a constant flow at speed $v_0$ in the $x$-direction, $\frac{dF}{dz} = v_0$ and hence $F(z) = v_0 ~z$ is the complex potential for such a constant fluid flow.
So that's the answer to your actual question, if the boundary conditions for your real flow problem are $\vec v(\vec r \rightarrow \infty) = v_0 \hat x$.