My teacher wrote this in the notes:
Let $f:D(a;R) \to \mathbb{C}$ be analytic, $z \in D(a;r)$ and define
$g:D(a;R) \to \mathbb{C}$ by $$ g(k)=\frac{f(k)-f(z)}{k-z} \ \ \
\text{when} \ k \neq z$$ $$ g(k)=f'(k) \ \ \ \text{when} \ k = z.$$
Then, $g$ is analytic on $D(a;R)\setminus\{z\}$ and continuous on $
D(a;r)$ . then, we have $$\int_{\partial D(a;r)}\frac{f(k)-f(z)}{k-z}
dk= \int_{\partial D(a;r)}\frac{f(k)}{k-z} dk-f(z)\int_{\partial
> D(a;r)}\frac{dk}{k-z}=0.$$ Direct computation shows that
$$\int_{\partial D(a;r)}\frac{dk}{k-z}=2\pi i.$$ Hence, we have the
Cauchy Integral Formula.
I don't see how 'direct' it is. I try to parameterize the circumference of the disk by $I(t)=a+re^{it}$, for $t \in [0, 2\pi]$. How should I proceed?
Best Answer
More details of this technique can be found in Griffiths and Harris.
Consider $$ \int_{\partial D(a;r)}\frac{dk}{k-z}. $$ We can rewrite it as $$ \int_{\partial D(a;r)}\frac{dk}{(k-a)-(z-a)}=\int_{\partial D(a;r)}\frac{dk}{(k-a)\left(1-\frac{z-a}{k-a}\right)}. $$ Now, we can use the power series expansion of $\frac{1}{1-x}$ to get that this is $$ \int_{\partial D(a;r)}\frac{dk}{k-a}\sum_{n=0}^\infty\left(\frac{z-a}{k-a}\right)^n. $$ Using absolute convergence of the sum (since $|z-a|<|k-a|$), you can interchange the sum and the integral to get $$ \int_{\partial D(a;r)}\frac{dk}{k-z}= \sum_{n=0}^\infty(z-a)^n\int_{\partial D(a;r)}\frac{dk}{(k-a)^{n+1}}. $$ Now, using the formula where $k=a+re^{it}$ and $dk=rie^{it}dt$, this becomes $$ \int_{\partial D(a;r)}\frac{dk}{k-z}= \sum_{n=0}^\infty(z-a)^n\int_0^{2\pi}\frac{rie^{it}dt}{r^{n+1}e^{(n+1)it}}. $$ Simplifying, we have $$ \int_{\partial D(a;r)}\frac{dk}{k-z}= \sum_{n=0}^\infty\frac{i(z-a)^n}{r^n}\int_0^{2\pi}e^{-nit}dt. $$ This is nonzero when $n=0$ because then the integral is $\int_0^{2\pi}dt=2\pi$ and the coefficient is $i$. This is zero when $n>0$ because then the integral is $$ \int_0^{2\pi}e^{-nit}dt=\left.-\frac{1}{ni}e^{-nit}\right|_0^{2\pi}=0. $$