= Solution
Set $f(z)=\cosh(\pi z/a)$. The <hyperbolic cosine> addition formula gives
$$
f(x+iy)=\cosh\frac{\pi x}a\cos\frac{\pi y}a+i\sinh\frac{\pi x}a\sin\frac{\pi y}a.
$$
For $x>0$ 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>:
$$
\begin{aligned}
y=0&:\quad f\in[1,\infty),\\
x=0&:\quad f\in[-1,1],\\
y=a&:\quad f\in(-\infty,-1].
\end{aligned}
$$
Also $f'(z)=(\pi/a)\sinh(\pi z/a)\ne0$ 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 $|s|>1$, $\operatorname{Im}s>0$, and $f(z)=(s+s^{-1})/2$. Given any $\operatorname{Im}\zeta>0$, 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 $|s|>1$, and
$$
\operatorname{Im}\zeta=\tfrac12(1-|s|^{-2})\operatorname{Im}s
$$
forces 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 $\operatorname{Re}c>0$ and $0<\operatorname{Im}c<a$. Put $C=f(c)$, so $\operatorname{Im}C>0$. The <method of images> places an equal <line source> at $\bar C$ below the <real axis>:
$$
W(\zeta)=\frac Q{2\pi}\left[\log(\zeta-C)+\log(\zeta-\bar C)\right].
$$
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>:
$$
\boxed{w(z)=\frac Q{2\pi}\left[\log\!\left(\cosh\frac{\pi z}a-\cosh\frac{\pi c}a\right)+\log\!\left(\cosh\frac{\pi z}a-\cosh\frac{\pi\bar c}a\right)\right].}
$$
This is the <line source in a closed-end semi-infinite strip>, with no extra imposed flow. Since $f'(c)\ne0$, near the physical source $f(z)-f(c)=f'(c)(z-c)+O((z-c)^2)$. 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> $Q$.
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