Integrate $\int_{|z|=1/2}\frac{e^{1-z}}{z^{3}(1-z)}dz$ verification

Question

I integrate over the edge of a circle $K$ with radius 1/2

$\int_{|z|=1/2}\frac{e^{1-z}}{z^{3}(1-z)}dz=\int_{|z|=1/2}-\frac{e^{1-z}}{z^{3}}\frac{1}{(z-1)}dz$

By the Cauchy Integral form

$f(w)=\frac{1}{2\pi i}\int_{\partial K}\frac{f(z)}{z-w}dw$

I get

$\int_{|z|=1/2}-\frac{e^{1-z}}{z^{3}}\frac{1}{(z-1)}dz=-2\pi i \frac{e^{1-1}}{1^{3}}=-2\pi i$

Is this the way to do it ? It seemed quite simple to me.

Edit: This exercise should be solved without the residue theorem.

Answer 1

Hint:$$\int_{\mid z\mid =1/2}\dfrac{\frac{e^{1-z}}{1-z}}{z^3}\mathrm dz=f^{(2)}(0)\cdot \dfrac{2\pi i}{2!}=\pi ie$$ $z=1$ doesn't lie inside the circle of radius $1/2$ centered at the origin of the Complex plane, so calculating the residue at $0$ is the way to go. Also you need to consider making use of the general Cauchy Integral formula stated as follows: $$f^{(k)}(w)=\dfrac{k!}{2\pi i}\int_{\gamma} \dfrac{f(z)}{(z-w)^{k+1}}\mathrm dz$$

Answer 2

$1$ is not within the circle $\mid z\mid=\frac12$. So instead, you need to calculate the residue at $0$. There's a formula: $\frac12\lim_{z\to0}\frac{\operatorname {d^2(z^3)f}}{\operatorname {dz^2}}$ in this case.

Answer 3

No, that is wrong. The singularity that you have to deal here is located at $0$. You can use the residue theorem:$$\int_{\lvert z\rvert=\frac12}\frac{e^{1-z}}{z^3(1-z)}=2\pi i\operatorname{res}_{z=0}\frac{e^{1-z}}{z^3(1-z)}=\pi ie.$$