Given $z = x + \mathrm{i} y$, we have:
\[ e^{\mathrm{i} z} = e^{\mathrm{i} (x + \mathrm{i} y)} = e^{\mathrm{i} x} e^{-y} \]
Using Euler's formula, $e^{\mathrm{i} x} = \cos x + \mathrm{i} \sin x$, we get:
\[ e^{\mathrm{i} z} = (\cos x + \mathrm{i} \sin x) e^{-y} =
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