Discussion
What is the period of \( e^{z} \) in the complex plane?
Answer: C
Since \( e^{z + 2\pi i} = e^{z}e^{2\pi i} \) and \( e^{2\pi i} = 1 \),
\[ e^{z + 2\pi i} \;=\; e^{z}. \]
So the exponential is periodic with period \( 2\pi i \), a purely imaginary period. This is the reason the complex logarithm is many-valued: the exponential is not one to one
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