Question
Obtain the Laurent series expansion of the function ez/(z -1)2 about z = 1, and determine the type of singularity.
Answer :
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Skip in 10
To find the Laurent series expansion of the function \( f(z) = \frac{e^z}{(z-1)^2} \) about \( z = 1 \), we can proceed step by step. ### Step 1: Express the function in terms of \( w = z - 1 \) Let \( w = z - 1 \), so that \( z = w + 1 \). The function becomes: \[ f(z) = \frac{e^z}{(z-1)^2} = \frac{e^{w+1}}{w^2} \] Now, expand \( e^{w+1} \) using the Taylor series expansion: \[ e^{w+1} = e \cdot e^w = e \left( 1 + w + \frac{w^2}{2!} + \frac{w^3}{3!} + \cdots \right) \]
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