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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 8.17,19,36 --> These exercises involve possibility of computing multiple Laurent series.

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(Q1) For Exer 8.17, do we obtain multiple Laurent series depending on how we rewrite $\frac1{z+1} = \frac1{2+z-1}$?

I did similarly for Exer 8.18, and it seems that that's the point based on Exer 8.32 where I obtained 4 Laurent series (the 4th being convergent on the $\emptyset$!)

For (Q1)

  1. Take out $z-1$ $$\frac1{z+1} = \frac1{2+z-1} = \frac1{(z-1)(\frac{2}{z-1}+1)} \to \ \text{a Laurent series for} \ |z-1| > 2$$

  2. Take out $2$

$$\frac1{z+1} = \frac1{2+z-1} = \frac1{(2)(1+\frac{z-1}{2})} \to \ \text{a Laurent series for} \ |z-1| < 2$$

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(Q2) For Exer 8.19, is there only one Laurent series unlike in Exer 8.17,18?

For (Q2)

$$\frac{z-2}{z+1} = 1 + \frac{-3}{z+1} \ \text{only}?$$

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(Q3) For Exer 8.36, I came up with 2 Laurent series. Are they both valid?

Rewrite $\frac1{(z^2-4)(z-2)} = \frac1{(z-2)^2(z+2)}$. Rewrite $\frac1{z+2} = \frac1{z-2+4}$:

  1. Take out $z-2$ $$\frac1{z+2} = \frac1{z-2+4} = \frac1{(z-2)(\frac{4}{z-2}+1)} \to \ \text{a Laurent series for} \ |z-2| > 4$$

--> This is not the book's answer, and it doesn't seem to have a $c_{-1}$. It looks like the integral will be 0.

(Q3.1) What's wrong with this Laurent series? I guess something like $C[2,1] \subsetneq \{|z-2| > 4\}$, so it doesn't apply or something.

  1. Take out $4$

$$\frac1{z+2} = \frac1{z-2+4} = \frac1{(4)(1+\frac{z-2}{4})} \to \ \text{a Laurent series for} \ |z-2| < 4$$

--> This is the book's answer (apart from the region), and it gives the same answer as with Cauchy Integral Formula 5.1 (and later Residue Theorem 9.10) namely $\frac{- \pi i}{8}$

(Q3.2) Book says that for $\frac1{(4)(1+\frac{z-2}{4})}$, the region of convergence is $\color{red}{0 <} |z-2| < 4$. Why $0 <$?

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Note that isolated singularities of a rational function specify different regions of convergence when expanding the function around a point in a Laurent series. So, for each of these different regions there is a specific representation of $f$ as Laurent series.

Ad Q.1: Yes, we obtain two different Laurent series expansions of $f$ around $z=1$, one for each region of convergence. Since there are simple poles at $z=1$ and $z=-1$ we have to distinguish two regions of convergence when expanding around the pole $z=1$. \begin{align*} D_1:&\quad 0< |z-1|<2\\ D_2:&\quad |z-1|>2 \end{align*}

  • The first region $D_1$ is a punctured disc with center $z=1$, radius $2$ and the pole at $z=-1$ at the boundary of the disc. It admits for the fraction with pole at $z=1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=-1$ a power series.

  • The region $D_2$ contains all points outside the disc with center $z=1$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.

Note that in $D_1$ we have to exclude $0$, since $f$ is not defined there.

We have in $D_1$ the representation \begin{align*} \frac{1}{(z-1)(z+1)}&=\frac{1}{z-1}\cdot\frac{1}{2+(z-1)}\\ &=\frac{1}{z-1}\cdot\frac{1}{2\left(1+\frac{z-1}{2}\right)}\\ &=\frac{1}{2}\cdot\frac{1}{z-1}\sum_{j=0}^\infty(-1)^j\left(\frac{z-1}{2}\right)^j\\ &=-\sum_{j=-1}^\infty\left(-\frac{1}{2}\right)^j(z-1)^j \end{align*} and in $D_2$ we have the representation \begin{align*} \frac{1}{(z-1)(z+1)}&=\frac{1}{z-1}\cdot\frac{1}{2+(z-1)}\\ &=\frac{1}{(z-1)^2}\cdot\frac{1}{1+\frac{2}{z-1}}\\ &=\frac{1}{(z-1)^2}\sum_{j=0}^\infty(-1)^j\left(\frac{2}{z-1}\right)^j\\ &=\frac{1}{4}\sum_{j=-\infty}^{-2}\left(-\frac{1}{2}\right)^j(z-1)^j \end{align*}

Ad Q.2: Yes, there is only one region of convergence, since we have only one simple pole at $z=-1$. The region of convergence is $|z-1|>0$ and the Laurent-series expansion is according to your calculation.

Ad Q.3,Q3.1: We obtain for $|z-2|>4$ the Laurent series expansion \begin{align*} \frac{1}{(z^2-4)(z-2)}&=\frac{1}{(z-2)^2}\cdot\frac{1}{z+2}\\ &=\frac{1}{(z-2)^3}\cdot\frac{1}{1+\frac{4}{z-2}}\\ &=\frac{1}{(z-2)^3}\sum_{j=0}^\infty(-4)^j\frac{1}{(z-2)^j}\tag{1}\\ &=\cdots \end{align*}

Ad Q.3.2: The function under consideration \begin{align*} \frac{1}{(z^2-4)(z-2)} \end{align*} has a pole at $z=2$. Since this function is not defined at $z=2$ we have to exclude $0$ from the region and get $0<|z-2|<4$.

Hint: This answer with some more detailed considerations might be helpful.

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  • $\begingroup$ Thanks Markus Scheuer! I'll analyse later. $\endgroup$ – BCLC Aug 7 '18 at 5:14
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    $\begingroup$ @BCLC: You're welcome. I've added a hint pointing to additional information. $\endgroup$ – Markus Scheuer Aug 7 '18 at 8:08

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