= Solution
First suppose $c_i\ne0$ and put $F=\widehat w/(U-c)$. The wall condition gives $F(\pm L)=0$. Substitution into the <Rayleigh equation for inviscid shear flow> gives
$$
[(U-c)^2F']'-k^2(U-c)^2F=0.
$$
Multiply by $\overline F$ and use <integration by parts>. The resulting <weighted identity for Howard's semicircle theorem> is
$$
\int_{-L}^L(U-c)^2Q\,dz=0,\qquad Q=|F'|^2+k^2|F|^2,\qquad I=\int Q\,dz>0.
$$
Its imaginary part gives $\int UQ=c_rI$, and its real part then gives $\int U^2Q=(c_r^2+c_i^2)I$. Set $a=U_{\min}$ and $b=U_{\max}$. Since $(U-a)(b-U)\ge0$,
$$
0\le\int(U-a)(b-U)Q\,dz
=\bigl[(a+b)c_r-ab-c_r^2-c_i^2\bigr]I.
$$
Completing the square proves <Howard's semicircle theorem>:
$$
\boxed{\left(c_r-\frac{U_{\max}+U_{\min}}2\right)^2+c_i^2\le\left(\frac{U_{\max}-U_{\min}}2\right)^2.}
$$
In particular every growing <normal mode> lies in the upper semicircle for $k>0$. If $c$ is real and outside $[a,b]$, $F$ is again regular, but the same identity has a strictly positive integrand unless the mode vanishes. Such a neutral mode is impossible. Consequently neutral regular modes also satisfy the bound; no division by $c_i$ is needed for this last conclusion. The nonreal proof avoids critical-level singularities altogether.
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