 Solution

Set $f(z)=\cosh (\pi z/a)$. The hyperbolic cosine addition formula givesFor $x>0$ and $0<y<a$, both factors in the imaginary part are positive. Thus the image lies in the upper half-plane. The three walls map onto the entire real axis:Also $f^{\prime }(z)=(\pi /a)\sinh (\pi z/a)\ne 0$ in the interior.

To verify that this is a one-to-one conformal map onto the whole upper half-plane, factor it through $s=e^{\pi z/a}$. This first map is one-to-one onto $|s|>1$, $\mathrm{Im}s>0$, and $f(z)=(s+s^{-1})/2$. Given any $\mathrm{Im}\zeta >0$, the roots of $s^2-2\zeta s+1=0$ have product one. Neither lies on $|s|=1$, since that would make $\zeta$ real. Exactly one root therefore has $|s|>1$, andforces that root into the upper half of the exterior disk. It has a unique preimage $z=(a/\pi )(\log |s|+i\arg s)$ with $0<\arg s<\pi$. This proves both surjectivity and injectivity. The construction is the hyperbolic-cosine map of a half-strip.

The source location $c$ is complex, with $\mathrm{Re}c>0$ and $0<\mathrm{Im}c<a$. Put $C=f(c)$, so $\mathrm{Im}C>0$. The method of images places an equal line source at $\bar{C}$ below the real axis:For real $\zeta$, the product of the two logarithm arguments is $|\zeta -C{|}^2>0$. Hence the streamfunction is constant along the wall, on locally consistent branches of the complex logarithm, and there is no normal velocity. Pulling back gives the requested complex potential of a two-dimensional potential flow:This is the line source in a closed-end semi-infinite strip, with no extra imposed flow. Since $f^{\prime }(c)\ne 0$, near the physical source $f(z)-f(c)=f^{\prime }(c)(z-c)+O((z-c{)}^2)$. Thus $w=(Q/(2\pi ))\log (z-c)+$ a regular term, verifying that the source strength remains $Q$ under the conformal map. The image source has no preimage inside the physical strip. Finally, as $x\to \infty$, $w(z)\sim Qz/a+$ a constant, so the far-field velocity is $(Q/a,0)$ and the channel carries exactly the injected volume flux $Q$.