MHB Yes, your solution is correct. Good job!

[aruwin]

Hello.
Can someone check if I got the answer right?

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

My solution:

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

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

 

[chisigma]

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

$\chi$ $\sigma$

 

[mathbalarka]

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 $2e^2$.