# Toronto Math Forum

## MAT334-2018F => MAT334--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 04:09:19 PM

Title: Q6 TUT 0202
Post by: Victor Ivrii on November 17, 2018, 04:09:19 PM
Find the Laurent series for the given function $f(z)$ about the indicated point. Also, give the residue of the function at the point.
$$f(z)=\frac{1}{e^z-1};\qquad z_0=0\quad \text{(four terms of the Laurent series)} .$$
Title: Re: Q6 TUT 0202
Post by: Meng Wu on November 17, 2018, 04:09:29 PM
$$\because e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots$$
\begin{align}\therefore e^z-1&=(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)-1\\&=z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots \end{align}
Added omitted calculation of order of pole: (thanks Chunjing Zhang for pointing it out)
$$g(z)=\frac{1}{f(z)}=\frac{1}{\frac{1}{e^z-1}}=e^z-1\\g(z_0=0)=e^0-1=0\\g'(z)=e^z \Rightarrow g(z_0=0)=e^0=1\neq 0 \\$$
Thus the order of the pole of $f(z)$ at $z_0=0$ is $1$.
Hence we let $$\frac{1}{e^z-1}=a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots$$
$$\therefore(e^z-1)(\frac{1}{e^z-1})=1\\\therefore (z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)(a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots)=1$$
$$\Rightarrow a_{-1}+a_0z+a_1z^2+a_2z^3+\frac{1}{2}z+\frac{a_0}{2}z^2+\frac{a_1}{2}z^3+\frac{a_{-1}}{6}z^2+\frac{a_0}{6}z^3+\frac{a_{-1}}{24}z^3+\cdots=1$$
$$\therefore \begin{cases}a_{-1}=1\\a_0z+\frac{a_{-1}}{2}z=0 \Rightarrow a_0+\frac{a_{-1}}{2}=0 \Rightarrow a_0=-\frac{1}{2}\\\frac{a_{-1}}{6}z^2+\frac{a_0}{2}z^2+a_1z^2=0 \Rightarrow \frac{a_{-1}}{6}+\frac{a_0}{2}+a_1=0 \Rightarrow a_1=\frac{1}{12}\\ \frac{a_{-1}}{24}z^3+\frac{a_0}{6}z^3+\frac{a_1}{2}z^3+a_2z^3 \Rightarrow \frac{a_{-1}}{24}+\frac{a_0}{6}+\frac{a_1}{2}+a_2=0 \Rightarrow a_2=0\end{cases}$$
Therefore, the first four terms of the Laurent series:
$$\require{cancel} \cancel{1+\frac{1}{z}+(-\frac{1}{2})z^2+(0)z^3} \\ \text{Typo correction: } \frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2$$
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The residue of given function at $z_0$ is the coefficient of $(z-z_0)^{-1}$, which is $1$.
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If $0(z^3)=0$ is not counted since its zero, we could have $a_3z^4+\frac{a_2}{2}z^4+\frac{a_1}{6}z^4+\frac{a_0}{24}z^4+\frac{a_{-1}}{120}z^4 \Rightarrow a_3=-\frac{1}{720}$
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Hence we have $\frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2-\frac{1}{720}z^3$.
Title: Re: Q6 TUT 0202
Post by: Heng Kan on November 17, 2018, 05:30:38 PM
I think you got the coefficients correctly but the Laurent series  is wrong.  Here is my answer. See the attatched scanned picture.
Title: Re: Q6 TUT 0202
Post by: Chunjing Zhang on November 17, 2018, 06:00:55 PM
I think maybe it is needed to first calculate the singularity of the original function, here pole of order 1, and then set the expansion start at power of -1.
Title: Re: Q6 TUT 0202
Post by: Victor Ivrii on November 28, 2018, 04:49:16 AM
We have two solutions based on two different ideas: undetermined coefficients and a clever  substitution of the power expansion into power expansion. Why clever? Because it chips out $1$ from $(e^z-1)/z$ rather than from $e^z-1$ (which would be an error)