Question

with the aid of series, show that the function f defined by means of the equations f (z) = { sin z/z, if z 6= 0 1, if z = 0 is entire. use that result to establish the limit limz→0 sin z z = 1.

Answer 1

Explanation:

To show that f is entire, we need to show that it is differentiable everywhere in the complex plane. For z = 0, we have f(z) = 1, which is clearly differentiable with derivative 0.

For z ≠ 0, we can use the power series expansion of sin z and write

sin z = z − z^3/3! + z^5/5! − z^7/7! + …

Dividing by z and taking the limit as z → 0, we get

limz→0 sin z/z = limz→0 (1 − z^2/3! + z^4/5! − z^6/7! + …) = 1

since all the terms in the parentheses approach 1 as z → 0.

Therefore, we have shown that f is differentiable everywhere in the complex plane, and hence is an entire function.