Using the Cauchy integral formula show that $\oint_{ |z|=2} \dfrac { e^zdz}{(z-1)^2(z-3)}= -\dfrac 3 2je\pi$

Question

Using the Cauchy integral formula show that $$\oint_{ |z|=2} \dfrac { e^zdz}{(z-1)^2(z-3)}= -\frac 3 2je\pi.$$

My work is in this picture. The answer is $-\frac 3 2je\pi$. I used the singularities and since $3$ wasn't included in the contour $|z|=2$, I omitted it. I don't know why I am getting a different answer.

I used partial fractions and had $$ \dfrac a{z-1} + \dfrac b{z+1}$$ I took the limit $$\lim_{ z \to 1} \dfrac {e^z}{z+1} = \dfrac e 2$$ and for $b$, I did $$\lim_{ z \to -1} \dfrac { e^z }{ z-1} = -\dfrac 1 {2e}$$ My $f(z)$ became $$\dfrac e{2(z-1)} - \dfrac 1{2e(z+1)}$$ I applied the Cauchy integral formula and I got $$\dfrac e2(2j\pi) - \dfrac 1{2e}(2j\pi) $$ but I need $$-\frac 3 2je\pi$$

Answer 1

You can do your calculation using your approach but your partial fraction decomposition is not correct.

You have

$$\frac 1{(z-1)^2(z-3)}= -\frac 14 \frac 1{z - 1}-\frac 12 \frac 1{(z - 1)^2}+ \frac 14 \frac 1{z - 3}$$

Now, you can use Cauchy's formula directly omitting the on $|z|\leq 2$ holomorphic part belonging to $\frac 1{z-3}$:

\begin{eqnarray*}\oint_{|z|=2}\frac{e^z}{(z-1)^2(z-3)}dz & = & -\frac 14 \oint_{|z|=2}\frac{e^z}{z - 1}dz-\frac 12 \oint_{|z|=2}\frac{e^z}{(z - 1)^2}dz \\ & = & -\frac 14\cdot 2\pi i \cdot e^1 -\frac 12 \cdot 2\pi i \cdot e^1 \\ & = & -\frac 32 \pi i e \end{eqnarray*}

Answer 2

The function $f(z)=e^z/(z-3)$ is analytic on the open disc $D=\{z:|z|<5/2\}$ which includes the circle $C=\{z:|z|=2\}$ which surrounds the point $z=1.$ So $$\int_C\frac {f(z)}{(z-1)^2}dz=2\pi i f''(1).$$

For $z\ne 3$ let $g(z)=e^z$ and $h(z)=1/(z-3).$ We have $f''(1)=g''(1)h(1)+2g'(1)h'(1)+g(1)h''(1)=e(\,h(1)+2h'(1)+h''(1)\,)$.... etc.