Discussion
Let \(f(z)\) be an analytic function, where \(z = x + iy\). If the real part of \(f(z)\) is \(\cosh x\cos y\), and the imaginary part of \(f(z)\) is zero for \(y = 0\), then \(f(z)\) is
Answer: D
The function is \(f(z) = \cosh z\).
Expand \(\cosh z\) with \(z = x + iy\):
\[ \cosh(x + iy) = \cosh x\cos y + i\sinh x\sin y \]
Its real part is \(\cosh x\cos y\), as required, and its imaginary part \(\sinh x\sin y\) is zero when \(y = 0\). Since \(\cosh z\) is analytic and an analytic function is fixed by its real part up to an imaginary constant, which the condition at \(y = 0\) sets to zero, this is the function.
\(\cosh x\,e^{-iy}\) has the right real part but is not analytic (it fails the Cauchy-Riemann equations). \(\cosh z\cos y\) mixes z with y and is not analytic, and \(\cosh z\,e^{z}\) has a different real part.
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