Convergence of $f_n(z) = \sin(z/n)$

Question

Show that $ f_n(z) = \sin(z/n) $for $z\in \mathbb{C} , n \in \mathbb{C}$ converges pointwise on the complex plane.

Then for $p>0 $show $\{f_n\}$ converges uniformly on $\{z : |z| \le p\}$ but does not converege uniformly on the complex plane.

Any hints would be greatly appreciated!

Answer 1

For pointwise convergence, you have to show that, for every value of $z\in\mathbb C$, the limit $$\lim_{n\to\infty} \sin\left(\frac zn\right)$$ exists. Given that:

1. The limit $$\lim_{n\to\infty}\frac zn$$ exists
2. The function $w\mapsto \sin w$ is continuous

this should be easy.

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To prove uniform convergence, you first need to know what the limit function is, and I will gladly help you with the next steps once you find out what the pointwise limit is.

For now, just two hints:

• the uniform limit, if it exists, is the same as the pointwise limit.
• if $|z|$ is small, then $\sin |z|$ is also small.