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The residue of e^z/z^3 from the pole-derivative formula

Example

At $0$ , the function $e^{z} / z^{3}$ has residue $1 / 2$ .

Facts & Assumptions

Given: The function $f (z) = e^{z} / z^{3}$ .

[L1]

For a pole of order $m$ , the residue is $\frac{1}{(m - 1)!}$ times the $(m - 1)$ st derivative of $(z - a)^{m} f (z)$ at the pole (Residue formula for a pole of order m).

[L2]

The complex exponential is entire and equals its own derivative, so all derivatives of $e^{z}$ are again $e^{z}$ (The complex exponential is entire and its complex derivative is itself).

Verification

technique · direct

1.1

At $0$ , the factor $z^{3} f (z)$ is $e^{z}$ , so $f$ has a pole of order $3$ there.

givenalgebra

2.1

Applying [L1] with $m = 3$ gives $Res ⁣ (\frac{e^{z}}{z^{3}}, 0) = \frac{1}{2!} \frac{d^{2}}{d z^{2}} (e^{z}) ∣_{z = 0} .$

step 1.1L1

3.1

By [L2], the second derivative of $e^{z}$ is still $e^{z}$ , so the right-hand side of step 2.1 is $\frac{1}{2} e^{0} = \frac{1}{2}$ .

step 2.1L2algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jean-Baptiste Campesato, MAT334 course page and notes index (standard reference, not scraped)