limit from sides in complex function..

**transgalactic** · July 12th 2010, 10:31 AM

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

**Also sprach Zarathustra** · July 12th 2010, 10:35 AM

שלום

**Jose27** · July 12th 2010, 10:48 AM

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 $z^2$ has a finite limit, and $\sin \left( \frac{z}{z+1} \right)$ has an essential singularity at $-1$ then $f$ has an essential singularity at $-1$ , 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 $z^{-1}$ . 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.

**transgalactic** · July 12th 2010, 10:54 AM

wrong place

**Jose27** · July 12th 2010, 11:07 AM

Originally Posted by transgalactic

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

Taking the limits [Math]0^+[/tex] and $0^-$ 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 $-\infty$ as a limit of a complex function?

**transgalactic** · July 12th 2010, 11:13 AM

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

**Jose27** · July 12th 2010, 11:23 AM

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 $z+1$ there, which blow up near $-1$ whereas the $z$ 's remain bounded.

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

**transgalactic** · July 12th 2010, 11:30 AM

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

**Jose27** · July 12th 2010, 11:40 AM

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 $z_0\in U\subseteq \mathbb{C}$ with $U$ open, $f,g:U\setminus \{z_0\} \rightarrow \mathbb{C}$ holomorphic and $f$ has a removable singularity at $z_0$ and $g$ has an essential singularity at $z_0$ then $fg$ has an essential singularity at $z_0$

**transgalactic** · July 12th 2010, 11:45 AM

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

**Jose27** · July 12th 2010, 11:49 AM

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

**transgalactic** · July 12th 2010, 11:54 AM

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?

**Jose27** · July 12th 2010, 12:07 PM

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?

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

**transgalactic** · July 12th 2010, 12:42 PM

Originally Posted by Jose27

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

i understand why
$\sin \left( \frac{1}{z} \right)$ has an essential singularity at $0$

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

**Jose27** · July 12th 2010, 01:16 PM

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

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