Discussion
Consider \(f(z) = e^{z}\), where \(z = x + iy\) and \(i = \sqrt{-1}\). Which of the following statements is correct ?
Answer: A
The correct statement is \(f\) is periodic.
\(e^{z + 2\pi i} = e^{z}e^{2\pi i} = e^{z}\), so \(e^{z}\) is periodic with the purely imaginary period \(2\pi i\), even though the real exponential \(e^{x}\) is not periodic.
Writing \(e^{z} = e^{x}(\cos y + i\sin y)\) settles the others. The modulus is \(|f| = e^{x}\), which equals 1 only on the imaginary axis. The argument is \(y + 2n\pi\), not \(y \pm n\pi\): adding an odd multiple of \(\pi\) would flip the sign of f. And since f is periodic, 'f is not periodic' is false.
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