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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.

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1 Answer 1

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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 $z^2f(z)$.

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  • $\begingroup$ 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? $\endgroup$
    – AlvinL
    Apr 6, 2016 at 16:15
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    $\begingroup$ @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, 2016 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 $a=\infty$? $\endgroup$
    – AlvinL
    Apr 6, 2016 at 16:22
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    $\begingroup$ @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, 2016 at 16:25
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    $\begingroup$ @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, 2016 at 16:25

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