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Thread: continuity in C

  1. #1
    Super Member Showcase_22's Avatar
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    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 $\displaystyle z \in \mathbb{C}$.

    Decide whether f is continous at 0 and explain your answer.

    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!
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  2. #2
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    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 $\displaystyle z=0$.
    Last edited by Jose27; Nov 18th 2009 at 12:49 PM.
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  3. #3
    Super Member Showcase_22's Avatar
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    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!
    Last edited by Showcase_22; Nov 19th 2009 at 01:38 AM.
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