Discussion
What kind of singularity does \( \dfrac{\sin z}{z} \) have at the origin?
Answer: B
Expanding the numerator and dividing,
\[ \frac{\sin z}{z} \;=\; \frac{z - z^{3}/6 + \cdots}{z} \;=\; 1 - \frac{z^{2}}{6} + \cdots, \]
a series with no negative powers at all. The limit is one, so assigning that value at the origin makes the function analytic there. The zero of the numerator cancels the zero of the denominator
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