I've been trying to do this, but I really don't understand how contour integration works. The variable $z=a+ib$ is complex.
\begin{equation} \oint_C \frac{z\,e^{z\lambda}\,dz}{(z+1)^3} \end{equation} For the contour $C$ : $|z-\frac{1}{2}| = 2 $
Do I have to find the poles first, right?
On
Cauchy's Integral Theorem says: Let $C$ be a simply connected closed positively oriented contour. If $f$ is analytic in some domain containing $C$ and $z_0\in C,$ then $$f(z_0)=\frac{1}{2\pi i}\int_C\frac{f(z)}{z-z_0}dz \tag 1$$
Our contour $C$ is defined as $$C:=\{z\in\mathbb{C}:|z-1/2|=2\}$$ i.e. the circle centered at $\left(\frac{1}{2},0\right)$ of radius $2$.
Note that $$\frac{ze^{\lambda z}}{(z+1)^3}=\frac{ze^{\lambda z}}{(z-(-1))^3}$$ is analytic everywhere in $C$ except at $z_0=-1$ since $z_0\in C$. Since we have a pole of order $3$, differentiate $(1)$ twice so that: $$f''(z_0)=\frac{2!}{2\pi i}\int_C\frac{f(z)}{(z-z_0)^3}dz=\frac{1}{\pi i}\int_C\frac{ze^{\lambda z}}{(z-(-1))^3}dz\implies\tag2$$ $$\int_C\frac{ze^{\lambda z}}{(z-(-1))^3}dz=\pi if''(z_0)\tag3 $$
where $f(z)=ze^{\lambda z}$ and $z_0=-1$.
Compute successive derivatives: $$f(z)=ze^{\lambda z}$$ $$f'(z)=e^{\lambda z}+\lambda ze^{\lambda z}=e^{\lambda z}(1+\lambda z)$$ $$f''(z)=\lambda e^{\lambda z}+\lambda e^{\lambda z}(1+\lambda z)=\lambda e^{\lambda z}(2+\lambda z)$$ Finally we have $$\int_C\frac{ze^{\lambda z}}{(z+1)^3}dz=\pi i\left(\lambda e^{-\lambda}(2-\lambda)\right)$$
In lines $(2)$ and $(3)$, I'm using the result that $$\int_C\frac{f(z)}{(z-z_0)^m}dz=\frac{2\pi i f^{(m-1)}(z_0)}{(m-1)!}\space\space\text{for $z_0\in C$ and $m\in\mathbb{Z}^+$}$$
Alternatively, you can use the straight-forward expression for the residue of a function at a pole of order $n$
$${\rm Res}\left(f\left(z\right),z_{0}\right)=\dfrac{1}{\left(n-1\right)!}\dfrac{{\rm d}^{n-1}}{{\rm d}z^{n-1}}\lim_{z\rightarrow z_{0}}\Bigg[\left(z-z_{0}\right)^{n}f\left(z\right)\Bigg]$$
In your case $f\left(z\right)=\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}}$, thus $z_{0}=-1$, $n=3$ and so
$${\rm Res}\left(\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}},-1\right)=\dfrac{1}{2!}\dfrac{{\rm d}^{2}}{{\rm d}z^{2}}\lim_{z\rightarrow -1}\Bigg[\left(z+1\right)^{3}\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}}\Bigg]=\frac{1}{2}\lambda e^{-\lambda}\left(2-\lambda\right)$$
You then get from the residue theorem that
$$\oint_{C\left(\frac{1}{2},\sqrt{2}\right)}\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}}{\rm d}z=2\pi i{\rm Res}\left(\dfrac{ze^{\lambda z}}{\left(z+1\right)^{3}},-1\right)=\pi i\lambda e^{-\lambda}\left(2-\lambda\right)$$
exactly as @coreyman317 got.