Show that $2f'(0)=\frac{1}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)-f(-\zeta)}{\zeta^{2}} d\zeta,$ whenever $0<r<1.$ Suppose $f: \mathbb{D} \rightarrow \mathbb{C}$ is holomorphic. Show that
$$2f'(0)=\frac{1}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)-f(-\zeta)}{\zeta^{2}} d\zeta,$$
whenever $0&lt;r&lt;1.$

So far my understanding of this problem is that I need to prove that we have $f(z)=\frac{f(z)-f(-z)}{2}$ for $z$ satisfying $|z|&lt;1$, since the rest is an application of Cauchy's Integral Formula. I noted that the result is true if an only if we have $f(z)=-f(-z)$, for all $z$ satisfying the said condition. Here is where I am stuck so far. Could you please provide a small hint to point me in the right direction, please provide no complete solution, only small hints.
By extension Cauchy's Integral Formula ,we have

$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)}{{(\zeta-z_0)}^{n+1}} d\zeta,$$

provided $f(\zeta) is \ analytic\ in\ |\zeta|\ = \ r$

So, 

$$2f'(0)=\frac{2*1!}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)}{\zeta^{2}} d\zeta$$
$$\quad =\frac{1}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)-(-\zeta)}{\zeta^{2}} d\zeta \quad (given)$$

And ,Since $f(\zeta)$  is  analytic in $|\zeta| = r $  ,
thus it is continuous and differentiable in that region,

So ,$2f(\zeta)$ and  $f(\zeta) - f(-\zeta)$ are two continuous and differentiable functions with same antiderivatives

So , Two functions must be equal.

Hence,$$2f(\zeta) = f(\zeta) - f(-\zeta)$$
$$f(\zeta) =  - f(-\zeta)$$
By extension Cauchy's Integral Formula ,we have

$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)}{{(\zeta-z_0)}^{n+1}} d\zeta,$$

provided $f(\zeta) is \ analytic\ in\ |\zeta|\ = \ r$

So, 

$$2f'(0)=\frac{2*1!}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)}{\zeta^{2}} d\zeta$$
$$\quad =\frac{1}{2\pi i}\int_{|\zeta|=r} \frac{f(\zeta)-(-\zeta)}{\zeta^{2}} d\zeta \quad (given)$$

And ,Since $f(\zeta)$  is  analytic in $|\zeta| = r $  ,
thus it is continuous and differentiable in that region,

So ,$2f(\zeta)$ and  $f(\zeta) - f(-\zeta)$ are two continuous and differentiable functions with same antiderivatives

So , Two functions must be equal.

Hence,$$2f(\zeta) = f(\zeta) - f(-\zeta)$$
$$f(\zeta) =  - f(-\zeta)$$