0
$\begingroup$

Let $$f(z) = \frac{z+1}{\sin ^2z}, a=0;$$ First we find the series for $\sin ^2z$: $$\sin^2 z = 1-\cos^2 z =\frac{1}{2}(1-\cos 2z)= \frac{1}{2}\left (1-\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(2z)^{2n}\right ) = \sum_{n=1}^\infty \frac{(-1)^n\cdot 4^n}{(2n)!}z^{2n}$$ which is valid in all of $\mathbb{C}$ because $\sin z$ is regular in $\mathbb{C}$ and the pointwise product (should be true for compositions, too) of regular functions is regular. So we have $$\frac{z}{\sin ^2z} + \frac{1}{\sin ^2z} \overset{?}= \frac{1}{\frac{\sin ^2z}{z}} +\frac{1}{\sin ^2z} = \frac{1}{\sum_{n=1}^\infty \frac{(-1)^n\cdot 4^n}{(2n)!}z^{2n-1}} + \frac{1}{\sum_{n=1}^\infty \frac{(-1)^n\cdot 4^n}{(2n)!}z^{2n}} $$

For the principal part I need something of the form $\sum_{k=-\infty}^{-1}\ldots$ and I seem to have made no progress whatsoever. I don't understand how I should get there.

The (bloody) textbook offers no examples about this problem, so I can't even tell if I'm on the right track.

$\endgroup$
1
$\begingroup$

Hint Note that $z^2 f(z)$ has a removable singularity at $z=0$. This means it has a Taylor series, and hence the Principal Part of your Laurent series must have the form $$\frac{b_1}{z}+\frac{b_2}{z^2}$$

to find this it is enough to find the first two terms from the Taylor series of $z^2f(z)$.

$\endgroup$
  • $\begingroup$ The removable singularity is due to $z \sim \sin z$ in the process $z\to 0$ as I understand. Because the singularity is removable, it means $z^2f(z)$ regular at $0$, but how'd you arrive at the conclusion on the form of the principal part? You simply divide by $z^2$ again and count the terms that have negative powers? $\endgroup$ – Alvin Lepik Apr 6 '16 at 16:15
  • 1
    $\begingroup$ @AlvinLepik $$z^2f(z)= \sum_{n=0}^\infty a_n z^n$$ What happens when you divide both sides by $z^2$? $\endgroup$ – N. S. Apr 6 '16 at 16:17
  • $\begingroup$ We get a series running from $\sum_{-2}^\infty$, I get it. Is this approach universal? How would one tackle a similar problem, perhaps for not the same function around $a=\infty$? $\endgroup$ – Alvin Lepik Apr 6 '16 at 16:22
  • 1
    $\begingroup$ @AlvinLepik The approach is universal in the following sense: if $f(z)=\frac{g(z)}{h(z)}$ with $g,h$ Analytic at $a$, then the principal part of the Laurent series of $f$ is finite (i.e. $a$ is a pole for $f$). The number of terms is simply the order of the zero $a$ in $h$ minus the order of the zero $a$ in $g$. $\endgroup$ – N. S. Apr 6 '16 at 16:25
  • 1
    $\begingroup$ @AlvinLepik Now, if $a=\infty$, just do a substitution $w=\frac{1}{z}$ to transform $a=\infty $ to $w=0$. $\endgroup$ – N. S. Apr 6 '16 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.