Math / Calculus

QuestionFind the remainder of the function $f(z)=\frac{e^{1 / z}}{(z-1)(z+2)^{2}}$ in the singular Point of a first-order pole.

Studdy Solution

STEP 1

1. The function $f(z) = \frac{e^{1/z}}{(z-1)(z+2)^2}$ is given.
2. We are looking for the residue at a first-order pole.
3. A first-order pole is a point where the function has a simple pole, i.e., the denominator has a factor of $(z-a)$ but not $(z-a)^2$ .

STEP 2

1. Identify the singular points of the function.
2. Determine which singular point is a first-order pole.
3. Calculate the residue at the first-order pole.

STEP 3

Identify the singular points of the function. These are the values of $z$ where the denominator is zero:
$(z-1)(z+2)^2 = 0$
The singular points are $z = 1$ and $z = -2$ .

STEP 4

Determine which singular point is a first-order pole. A first-order pole occurs when the factor in the denominator is linear, i.e., $(z-a)$ .
- At $z = 1$ , the factor is $(z-1)$ , which is linear, indicating a first-order pole. - At $z = -2$ , the factor is $(z+2)^2$ , which is quadratic, indicating a second-order pole.
Thus, the first-order pole is at $z = 1$ .

STEP 5

Calculate the residue at the first-order pole $z = 1$ . The residue at a simple pole $z = a$ of a function $g(z) = \frac{h(z)}{z-a}$ is given by:
$\text{Residue} = h(a)$
For $f(z) = \frac{e^{1/z}}{(z+2)^2}$ at $z = 1$ , we have:
$h(z) = \frac{e^{1/z}}{(z+2)^2}$
Evaluate $h(z)$ at $z = 1$ :
$h(1) = \frac{e^{1/1}}{(1+2)^2} = \frac{e}{9}$
Therefore, the residue at $z = 1$ is $\frac{e}{9}$ .
The remainder (residue) of the function at the first-order pole $z = 1$ is $\boxed{\frac{e}{9}}$ .

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