I'm currently trying to evaluate the following integrals:
$$\int_{|z-2|=1}\ {((e^z-1)^2/z) dz}$$
$$\int_{|z|=1}\ {((e^z-1)^2/(z^n))} dz$$ where $n$ belongs to positive, natural numbers.
I know that I should probably use Cauchy's Integral Formula or Cauhcy's Theorem, however I have a lot of difficulty understanding how and why I would use them to evaluate these integrals.
I would appreciated any help. Thanks :)
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Why you should use Cauchy's theorem: Because, if it is appliable, it helps you to calculate integrals very quickly. Let's do the first: $$ \int_{|z-2| = 1} \frac{(e^z - 1)^2}z\, dz $$ Recall that $f(z) = \frac{(e^z - 1)^2}z$ is holomorphic in $\mathbf C \setminus \{0\}$, especially in the simply connected domain $\{z \in \mathbf R: \Re z > 1\}$. As the integration path $z:|z-2|= 1$ lies in this simply connected domain, we have by Cauchy $$ \int_{|z-2| = 1} \frac{(e^z - 1)^2}z\, dz = 0 $$ tl;dr: As $0$, the only point, where the integrand is not holomorphic, lies outside of the integration path, by Cuachy, the integral in $0$.
This simple trick does not work in the second case, because $z = 0$ lies on the inner side of the contour $\{z: |z|=1\}$, so we have to do something else. Cauchy's formula reads $$ f^{(n)}(a) = \frac {n!}{2\pi i} \int_\gamma \frac{f(z)}{(z-a)^{n+1}}\, dz $$ We apply it here for $f(z) = (e^z - 1)^2$ and $n-1$ for $a = 0$, we have $$ f^{(n-1)}(0) = \frac{(n-1)!}{2\pi i } \int_{|z| = 1} \frac{(e^z - 1)^2}{z^n} \, dz $$ So $$ \int_{|z| = 1}\frac{(e^z - 1)^2}{z^n}\, dz = \frac{2\pi i }{(n-1)!} f^{(n-1)}(0) $$ As \begin{align*} f(z) &= (e^z - 1)^2 = e^{2z} - 2e^z + 1\\ f'(z) &= 2e^{2z} - 2e^z\\ f''(z) &= 4e^{2z} - 2e^z\\ &\vdots\\ f^{(n-1)}(z) &= 2^{n-1}e^{2z} - 2e^z, \qquad n \ge 2 \end{align*} Therefore $$ \int_{|z| = 1} \frac{(e^z - 1)^2}{z^n}\, dz = \frac{2\pi i}{(n-1)!} (2^{n-1} - 2), \quad n \ge 2 $$
For the first integral, note that the integrand is holomorphic on and in the disk $|z-2|=1$ because the singularity $z=0$ is outside it.
For the second integral, just expand the integrand in Laurent series, which is easy because you know the Taylor series for $e^z$.