Yes, your solution is correct. Good job!

In summary, the conversation discusses evaluating the residue of a complex function $f(z)=\frac{e^{-2z}}{(z+1)^2}$ at $z=-1$. The solution involves taking the limit of the derivative of the function and applying a formula for poles of order $n$, resulting in a residue of $2e^2$. The conversation also mentions a classic way to check the solution by evaluating the Laurent series of the function.
  • #1
aruwin
208
0
Hello.
Can someone check if I got the answer right?

$f(z)=\frac{e^{-2z}}{(z+1)^2}$

My solution:

$f(z)=\frac{e^{-2z}}{(z+1)^2}$
$$Resf(z)_{|z=-1|}=\lim_{{z}\to{-1}}\frac{d}{dz}((z+1)^2\frac{e^{-2z}}{(z+1)^2})$$

$$\lim_{{z}\to{-1}}-2e^{-2z}=-2e^{2}$$
 
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  • #2
aruwin said:
Hello.
Can someone check if I got the answer right?

$f(z)=\frac{e^{-2z}}{(z+1)^2}$

My solution:

$f(z)=\frac{e^{-2z}}{(z+1)^2}$
$$Resf(z)_{|z=-1|}=\lim_{{z}\to{-1}}\frac{d}{dz}((z+1)^2\frac{e^{-2z}}{(z+1)^2})$$

$$\lim_{{z}\to{-1}}-2e^{-2z}=-2e^{2}$$

You are right!... in general for a pole of order n z=a is...

$\displaystyle \text{Res}_{z=a}\ f(z) = \frac{1}{(n-1)!}\ \lim_{z \rightarrow a} \frac{d^{n-1}}{d z^{n-1}}\ \{(z-a)^{n}\ f(z)\}\ (1)$

Kind regards

$\chi$ $\sigma$
 
  • #3
There's always the classic way to do it/check your solution, though : We want to evaluate the Laurent series of

$$\frac{\exp(-2z)}{(1 + z)^2}$$

At $z = -1$. Substituting $1 + z = t$ gives

$$\frac{\exp(2 - 2t)}{t^2} = \frac{e^2}{t^2} \cdot \left ( 1 - 2 \cdot t + 2 t^2 + \mathcal{O}(t^3) \right )= \frac{e^2}{t^2} - \frac{2e^2}{t} + 2e^2 + \mathcal{O}(t) $$

Thus, the residue of the expression (which is the coefficient of $t^{-1}$ in the Laurent series) at $z = -1$ is $2e^2$.
 
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