# Thread: limit from sides in complex function..

1. ## limit from sides in complex function..

i was told here that there is no +infinity and -infinity in complex functions
only infinity

but i have a solved question from a test which is solved exactly like it you said was wrong
and got full credit

maybe i am missing something here

2. שלום

3. Originally Posted by transgalactic
i was told here that there is no +infinity and -infinity in complex functions
only infinity

but i have a solved question from a test which is solved exactly like it you said was wrong
and got full credit

maybe i am missing something here

Really, it would have sufficed to note that since $\displaystyle z^2$ has a finite limit, and $\displaystyle \sin \left( \frac{z}{z+1} \right)$ has an essential singularity at $\displaystyle -1$ then $\displaystyle f$ has an essential singularity at $\displaystyle -1$, then what you showed (basically) is that the limit doesn't exist. I assume they gave you the credit since your argument vaguely appeals to what I said.

As an example of what you're doing wrong take $\displaystyle z^{-1}$. Clearly we can think of it as a real or complex function. Analyze the behaviour in both cases and remember why we can talk about poles.

4. wrong place

5. Originally Posted by transgalactic
so you still say that 0+ and 0- +-infinity method here is wwrong?
Taking the limits [Math]0^+[/tex] and $\displaystyle 0^-$ is not wrong, you're just taking two different curves to approach the limit (which is always valid), what is wrong is your distinction of infinities. Tell me how would you define $\displaystyle -\infty$ as a limit of a complex function?

6. so you still say that 0+ and 0- +-infinity method here is wwrong?

and i did developed the function into a series

and i saw that there is no z in a negative power

so it cannot be significant singular point

7. Originally Posted by transgalactic
so you still say that 0+ and 0- +-infinity method here is wwrong?

and i did developed the function into a series

and i saw that there is no z in a negative power

so it cannot be significant singular point
What about the powers of $\displaystyle z+1$ there, which blow up near $\displaystyle -1$ whereas the $\displaystyle z$'s remain bounded.

By the way, please answer the question in my previous post, it will clarify things for you.

8. if my development is correct how whould you continue to develop it
so we will get a series with only one z variable

??
i cant make a guess here its all has to be series with only one z in a certain power in order to makea conclution

9. Frankly I wouldn't know how to turn that expression into a Laurent series, but it's not necessary, that is why one develops propositions like: if $\displaystyle z_0\in U\subseteq \mathbb{C}$ with $\displaystyle U$ open, $\displaystyle f,g:U\setminus \{z_0\} \rightarrow \mathbb{C}$ holomorphic and $\displaystyle f$ has a removable singularity at $\displaystyle z_0$ and $\displaystyle g$ has an essential singularity at $\displaystyle z_0$ then $\displaystyle fg$ has an essential singularity at $\displaystyle z_0$

10. what is f
what is g
in our case
?
could you explain your method more thurely
i dont know how to apply it here

11. $\displaystyle f(z)=z^2, \ g(z)=\sin \left( \frac{z}{z+1} \right), \ U=\mathbb{C} \setminus \{ -1 \}$

12. ok and how to prove that at -1 the sin expression is significant
you canceled my +-inf method
and by developing it into a series is not working either
so how to prove that at -1 the sin expression is significant?

13. Originally Posted by transgalactic
ok and how to prove that at -1 the sin expression is significant
you canceled my +-inf method
and by developing it into a series is not working either
so how to prove that at -1 the sin expression is significant?
$\displaystyle \sin (z)$ has an essential singularity at infinity, which means $\displaystyle \sin \left( \frac{1}{z} \right)$ has an essential singularity at $\displaystyle 0$ (this is easy to check since the principal part of the Laurent expansion has infinitely many terms) and this gives that $\displaystyle \sin ( f(z))$ has an essential sing. wherever $\displaystyle f$ has a pole. Take $\displaystyle f(z)= \frac{z}{z+1}$ and the conclusion follows.

14. Originally Posted by Jose27
$\displaystyle \sin (z)$ has an essential singularity at infinity, which means $\displaystyle \sin \left( \frac{1}{z} \right)$ has an essential singularity at $\displaystyle 0$ (this is easy to check since the principal part of the Laurent expansion has infinitely many terms) and this gives that $\displaystyle \sin ( f(z))$ has an essential sing. wherever $\displaystyle f$ has a pole. Take $\displaystyle f(z)= \frac{z}{z+1}$ and the conclusion follows.
i understand why
$\displaystyle \sin \left( \frac{1}{z} \right)$ has an essential singularity at $\displaystyle 0$

but how we can take any function we want and say sin(f) has significant point
at the pole of f
??
like here
$\displaystyle \sin ( f(z))$
$\displaystyle f(z)= \frac{z}{z+1}$

15. Basically, because near a pole, a function behaves like $\displaystyle \frac{1}{(z-z_0)^m}$ for some integer m. Try to work out the details by yourself (maybe using the multiplication formula for $\displaystyle \sin$ will simplify the argument). For this specific case a simple change of variables (and maybe a little geometric argument) gives the conclusion.