1. ## continuity in C

Let $\displaystyle f(z)= \begin{cases} z \cos \left( \frac{1}{z} \right) \ \ \ x \neq 0 \\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ x=0 \end{cases}$ where $\displaystyle z \in \mathbb{C}$.

I don't think $\displaystyle f(z)$ is continuous. However, I am having some trouble constructing two sequences that tend to 0 but have different limits when put into $\displaystyle f(z)$.

I think the sequences are going to be like $\displaystyle a_n=\frac{1}{2n \pi+ \frac{\pi}{2}}$ but they aren't working!

2. If $\displaystyle \lim_{z\rightarrow 0} z^k \cos ( \frac{1}{z} )$ exists then $\displaystyle \cos (\frac{1}{z} )$ has a pole of order $\displaystyle \leq k$. But take $\displaystyle a_k= \frac{1}{\pi /2+ \pi k}$ and $\displaystyle b_k= \frac{1}{2\pi k}$ then $\displaystyle \cos (\frac{1}{a_k} )=0$ and $\displaystyle \cos (\frac{1}{b_k} )=1$ which means $\displaystyle \lim_{z\rightarrow 0} \cos (\frac{1}{z} )$ does not exist ie. the function has an essential singularity at $\displaystyle z=0$.

3. Thanks but I managed to prove it another way!

I'm afraid my course hasn't covered what a "pole of order $\displaystyle \leq k$" is and what an essential singularity is.

I'll read up on those, thankyou!

EDIT: Those sequences are real but z is complex. I can prove that $\displaystyle \cos \left( \frac{1}{z} \right)$ is not continous quite easily, it's just that when there's a z in front it makes things a bit trickier (this is kind of what I mean):

$\displaystyle f(b_k)=b_k \cos \left( \frac{1}{b_k} \right)=0$

Similarly:

$\displaystyle f(a_k)=a_k \cos \left( \frac{1}{a_k} \right)=0$

Hence they have the same limit so in $\displaystyle \mathbb{R}, \ f(z)$ is continuous. To me this implies that you would need two sequences, one approaching 0 from the imaginary axis and one approaching 0 from the real axis that have the same limit (ie. 0) but f(sequence 1) $\displaystyle \neq$ f(sequence 2) (when limits are applied). This is how I did it but it's about a page long. What you've done is shorter!