Calculate the integrals below using the theory seen on Cauchy's integral formula. Do not use the formula that involves parameterization.
a)$\int_{\gamma} \frac{e^{iz}}{z^2}dz$, where $\gamma(t) = e^{it}, 0 \le t \le 2\pi$.
b)$\int_{\gamma} \frac{log z}{(z-1)^n}dz$, where $\gamma(t) = 1 + \frac{1}{2}e^{it} , 0 \le t \le 2\pi$ e $n \in \mathbb N$.
a)Let $a=0,n=1,f(z) = e^{iz} \to f'(z) = ie^{iz}$ so $f(z)$ is analitic on $\mathbb C$ and $B(0,1) \subset \mathbb C$. From
$$ f^{n(n)}(a) = \frac{n!}{2 \pi i} \int_{\gamma} \frac{f(w)}{(w-a)^{n+1}}dw$$ where $\gamma (t) = a + re^{it}, 0 \le t \le 2 \pi$.
We have
$f'(0) = \frac{1!}{2 \pi i} \int_{\gamma} \frac{e^{iz}}{(z-0)^2} dz \to i = \frac{1}{2 \pi i} \int_{\gamma} \frac{e^{iz}}{z^2} dz \to \int_{\gamma} \frac{e^{iz}}{z^2} dz = - 2 \pi$.
b)Let $a =-1, n=n,f(z)=log z \to f'(1)= \frac{1}{zln10} \to f^{2}= - \frac{1}{z^2 ln 10} \to f^{3}= \frac{2}{z^3 ln 10} \to f^{4}= - \frac{6}{z^4 ln 10} \to f^{5}= \frac{24}{z^5 ln 10}$
I think a) is correct but I'm stuck on b).
Thanks.
Cauchy integral formula says:
$\oint \frac {f(z)}{z-a}^n \ dz = 2\pi i \frac {f^{(n-1)}(a)}{(n-1)!}$
But, calculating those derivatives can be a pain in the butt. Here is a shortcut.
Remember that that the Taylor series of a function is:
$f(x) = f(a) + f'(a)(z-a) + \cdots +\frac {f^{(n)}(a)}{n!}(z-a)^n + \cdots$
Those coefficients are exactly what we are looking for. If we know the Talor series of our function, all we have to do is find the right coefficient.
$\log z = \sum_\limits{n=1}^\infty \frac {(-1)^{n+1}}{n} (z-1)^n$
(in complex analysis and every math class thereafter "log" should be assumed to be $\ln$)
When $n>1$
$\oint \frac {\log z}{(z-1)^n} \ dz = 2\pi i\frac {(-1)^{n}}{n-1}$