Let f be holomorphic on a region containing the closed unit disk $D(0, 1)$ and C be the unit circle traversed anticlockwise. Let n be a positive integer. Show that $$f(0) + f'(0)=\frac{1}{2\pi i} \int_C\frac{f(w)e^w}{w^2}dw$$ and that $$\frac{1}{2\pi i} \int_C\frac{f(w)+f(e^{2\pi i/n}w)+...+f(e^{2(n-1)\pi i/n}w)}{w^2}dw=0$$
I am not allowed to use the residue theorem.
I know that $f(0)=\frac{1}{2\pi i} \int_C\frac{f(w)}{w}dw$ and $f'(0)=\frac{1}{2\pi i} \int_C\frac{f(w)}{w^2}dw$ but that only gets me to $f(0) + f'(0)=\frac{1}{2\pi i} \int_C\frac{f(w)(w+1)}{w^2}dw$.
For the second part I tried writing every term within the integral as $\frac{f(e^{2(n-1)\pi i/n}w)}{w^2 {(e^{2(n-1)\pi i/n})}^2}*{(e^{2(n-1)\pi i/n})}^2$ but to no avail as I get to $f'(0) + ... + \frac{f'(0)}{{(e^{2(n-1)\pi i/n})^2}}$.
Any hints/tips would be greatly appreciated. Thanks.
If $n > 1$, then the second part may be proven as follows. Let $\lambda = e^{2\pi i/n}$. For $0\le k \le n-1$, the Cauchy integral formula for derivatives yields
$$\frac{d}{dw}\bigg|_{w = 0} f(\lambda^k w) = \frac{1}{2\pi i} \int_C \frac{f(\lambda^k w)}{w^2}\, dw.$$
By the chain rule, $\frac{d}{dw}\bigg|_{w = 0} f(\lambda^k w) = \lambda^k f'(0)$. Thus
$$\lambda^k f'(0) = \frac{1}{2\pi i} \int_C \frac{f(\lambda^k w)}{w^2}\, dw\qquad (k = 0,1,\ldots, n-1).$$
Summing the equations from $k = 0$ to $n-1$ results in
$$(1 + \lambda + \cdots + \lambda^{n-1})f'(0) = \frac{1}{2\pi i} \int_C \frac{f(w) + f(\lambda w) + \cdots + f(\lambda^{n-1} w)}{w^2}\, dw\tag{*}$$
Since $n > 1$, $\lambda \neq 1$ with $\lambda^n = 1$. So the left-hand side of (*) equals
$$ \frac{1-\lambda^n}{1 - \lambda}f'(0) = 0 f'(0) = 0,$$
and the result follows.