the residu of f(z) = 1 - e^-z divided by z^3 at z=0

To find the residue of the function 

\[
f(z) = \frac{1 - e^{-z}}{z^3}
\]

at \( z = 0 \), we first need to expand the function in a Laurent series around \( z = 0 \).

### Step-by-Step Solution:

1. **Expand the Exponential Function**:

   The exponential function \( e^{-z} \) can be expanded as a Taylor series around \( z = 0 \):

   \[
   e^{-z} = 1 - z + \frac{z^2}{2!} - \frac{z^3}{3!} + \cdots
   \]

   Thus, 

   \[
   1 - e^{-z} = z - \frac{z^2}{2!} + \frac{z^3}{3!} - \cdots
   \]

2. **Divide by \( z^3 \) to Find the Laurent Series**:

   Now, consider the function \( f(z) = \frac{1 - e^{-z}}{z^3} \):

   \[
   f(z) = \frac{z - \frac{z^2}{2!} + \frac{z^3}{3!} - \cdots}{z^3} = \frac{z}{z^3} - \frac{\frac{z^2}{2!}}{z^3} + \frac{\frac{z^3}{3!}}{z^3} - \cdots
   \]

   Simplify each term:

   \[
   f(z) = \frac{1}{z^2} - \frac{1}{2z} + \frac{1}{6} - \frac{z}{24} + \cdots
   \]

   The Laurent series for \( f(z) \) around \( z = 0 \) is:

   \[
   f(z) = \frac{1}{z^2} - \frac{1}{2z} + \left( \frac{1}{6} \right) - \frac{z}{24} + \cdots
   \]

3. **Identify the Coefficient of \( \frac{1}{z} \) (Residue Term)**:

   The residue of \( f(z) \) at \( z = 0 \) is the coefficient of \( \frac{1}{z} \) in its Laurent series. Here, the coefficient of \( \frac{1}{z} \) is:

   \[
   -\frac{1}{2}
   \]

### Conclusion:

The residue of the function \( f(z) = \frac{1 - e^{-z}}{z^3} \) at \( z = 0 \) is:

\[
\text{Res}(f, 0) = -\frac{1}{2}.
\]

Would you like more details on this, or do you have any other questions?

### Related Questions:

1. How would you find the residue of \( f(z) = \frac{e^z - 1}{z^4} \) at \( z = 0 \)?
2. What is the general method for finding residues of higher-order poles?
3. How can we use Laurent series to compute integrals in complex analysis?
4. What is the significance of the residue in complex integration (Cauchy's residue theorem)?
5. How does the Laurent series compare to the Taylor series in complex analysis?

**Tip:** Remember that the residue of a function at a point can be very useful in evaluating certain types of integrals using the residue theorem.

Learn how to calculate the residue of the function f(z) = (1 - e^{-z}) / z^3 at z = 0 using the Laurent series expansion method and Cauchy's residue theorem. This problem delves into complex analysis concepts, including the calculation of residues and Laurent series in complex functions. Suitable for advanced undergraduate students and beyond.

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