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
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.
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?
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 $\displaystyle z_0$
$\displaystyle \sin (z)$ 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
$\displaystyle \sin ( f(z))$
$\displaystyle f(z)= \frac{z}{z+1}$
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.