Find the Laurent series of all z 1? e² 2-1 about z 1. Why is this Laurent expansion valid for
Find the Laurent series of all z 1? e² 2-1 about z 1. Why is this Laurent expansion valid for
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Find the Laurent series of \(\frac{e^z}{z-1}\) about \(z = 1\). Why is this Laurent expansion valid for all \(z \neq 1\)?
**Solution:**
To find a Laurent series for \(\frac{e^z}{z-1}\) at \(z = 1\), we first substitute \(z = 1 + w\), so \(z - 1 = w\). The function becomes:
\[
\frac{e^{1+w}}{w} = \frac{e \cdot e^w}{w}
\]
This can be expanded using the Taylor series for \(e^w = 1 + w + \frac{w^2}{2!} + \frac{w^3}{3!} + \cdots\).
Thus,
\[
\frac{e}{w} \left(1 + w + \frac{w^2}{2!} + \frac{w^3}{3!} + \cdots\right) = \frac{e}{w} + e + \frac{e \cdot w}{2!} + \frac{e \cdot w^2}{3!} + \cdots
\]
This expression is a valid Laurent series for all \(z \neq 1\) because it accounts for the singularity at \(z = 1\) using negative powers in the expansion. Here, the term \(\frac{e}{w}\) equals \(\frac{e}{z-1}\), demonstrating the principal part of the Laurent series, confirming validity outside the singular point.
**Conclusion:**
Hence, the Laurent series of \(\frac{e^z}{z-1}\) about \(z = 1\) is valid for all \(z \neq 1\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa62f7b35-7db6-46d8-92c3-a45ad2747ea7%2F5b442736-208c-403e-88c3-1506367aa553%2Fw85mhql_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the Laurent series of \(\frac{e^z}{z-1}\) about \(z = 1\). Why is this Laurent expansion valid for all \(z \neq 1\)?
**Solution:**
To find a Laurent series for \(\frac{e^z}{z-1}\) at \(z = 1\), we first substitute \(z = 1 + w\), so \(z - 1 = w\). The function becomes:
\[
\frac{e^{1+w}}{w} = \frac{e \cdot e^w}{w}
\]
This can be expanded using the Taylor series for \(e^w = 1 + w + \frac{w^2}{2!} + \frac{w^3}{3!} + \cdots\).
Thus,
\[
\frac{e}{w} \left(1 + w + \frac{w^2}{2!} + \frac{w^3}{3!} + \cdots\right) = \frac{e}{w} + e + \frac{e \cdot w}{2!} + \frac{e \cdot w^2}{3!} + \cdots
\]
This expression is a valid Laurent series for all \(z \neq 1\) because it accounts for the singularity at \(z = 1\) using negative powers in the expansion. Here, the term \(\frac{e}{w}\) equals \(\frac{e}{z-1}\), demonstrating the principal part of the Laurent series, confirming validity outside the singular point.
**Conclusion:**
Hence, the Laurent series of \(\frac{e^z}{z-1}\) about \(z = 1\) is valid for all \(z \neq 1\).
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