Evaluating complex Integral over a disc

Question

I have the integral: $$\int _{\partial D(a, r)} \frac{e^z}{z^3 + 2z^2 + z} dz$$

which I have to find for different cases:

1 - $a = 0$ and $r =1/2$

2 - $a = -i - 1$ and $r = 1/2$

3 - $a = -1$ and $r = 1/2$

4 - $a = 0$ and $r = 2$

My attempt is this:

$$ \frac{e^z}{z^3 + 2z^2 + z} = \frac{e^z}{z(z+1)^2}$$

let $f(z) =\large \frac{e^z}{(z+1)^2}$, then the integral becomes:

$$\int _{\partial D(a, r)} \frac{f(z)}{z - 0} dz$$

And by Cauchy's equation this is equal to $2 \pi i f(0)$. $f(0)$ is $1$ in this case so the integral equals $2 \pi i$.

What i don't understand is where should i use the properties of the disc? Unless my approach is wrong

Any help would be awesome.

Answer 1

if $a=0$ and $r=2$, look the $f(z) = \frac{e^z}{(z+1)^2}$, it has singularity $z=-1$ in $D(a,r)$, so can not use Cauchy formula.

Answer 2

For $2$) observe that $f(z)=\frac{e^z}{z(z+1)^2}$ is analytic in the disc $D(a,r)$, then from the Cauchy's Theorem we get $$\int_{\partial D(a,r)}\frac{e^z}{z(z+1)^2}\,dz=0.$$

In the case $3$) we can take the function $g(z)=\frac{e^z}{z}$, which is analytic in the disc $D(-1,\frac12)$. Then from the Cauchy's integral formula we get $$\int_{\partial D(-1,\frac12)}\frac{e^z/z}{(z+1)^2}\,dz=2\pi i\frac{g'(-1)}{1!}=\left.2\pi i\cdot\frac{e^z(z-1)}{z^2}\right|_{z=-1}=-4\pi ie^{-1}$$

Answer 3

For (1) when $|z|<\dfrac12$ we have $$\int_{|z|<\frac12}\frac{e^z}{z(z+1)^2}\,dz=\int_{|z|<\frac12}\dfrac{\frac{e^z}{(z+1)^2}}{z}\,dz=2\pi i\frac{e^z}{(z+1)^2}\Big|_{z=0}=\color{blue}{2\pi i}$$