Discussion
For the value of \(\alpha\) obtained in Q.54, the function \(g(z)\) is not conformal at a point
Answer: A
\(g\) is not conformal at \(\frac{\pi(1 + 3i)}{6}\).
With \(\alpha = 3\) from Q.54, \(g(z) = \sinh(z - \pi/6)\). An analytic map fails to be conformal exactly where its derivative vanishes:
\[ \begin{aligned} g'(z) &= \cosh(z - \pi/6) = 0 \\ z - \pi/6 &= i\left(\tfrac{\pi}{2} + k\pi\right) \end{aligned} \]
since \(\cosh w = \cos(iw)\) vanishes only at \(w = i(\pi/2 + k\pi)\). With \(k = 0\), \(z = \frac{\pi}{6} + \frac{i\pi}{2} = \frac{\pi(1 + 3i)}{6}\).
\(\frac{\pi(3 + i)}{6}\) swaps the real and imaginary parts, and \(\frac{2\pi}{3}\) is real, where \(\cosh\) is never zero. \(\frac{i\pi}{2}\) is the zero of \(\cosh z\) itself, forgetting the shift by \(\pi/6\).
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