Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 331 / 1 / b / i / Solution

For a possibly unstable normal mode, $c_i\ne 0$ ensures $V=U-c$ never vanishes, so choose one continuous branch of $V^a$. Substitute $\hat{w}=V^{a}q$ into the Taylor–Goldstein equation and collect terms:Multiplication by $V^{2a}$ puts its first two terms in divergence form:Multiply by the complex conjugate $\overline{q}$ and apply integration by parts. Since $q(\pm L)=0$, the boundary term vanishes and the power-transformed Taylor–Goldstein energy identity isThe weight is $V^{2a}$, not $|V{|}^{2a}$: preserving its complex phase is essential for the subsequent stability proofs.