$2f'(0) = \frac{1}{2\pi i} \int_{|z|=r}\frac{f(z)-f(-z)}{z^2} dz$

Question

Suppose $f$ is a holomorphic function on a unit disc. Then how can we show that

$2f'(0) = \frac{1}{2\pi i} \int_{|z|=r}\frac{f(z)-f(-z)}{z^2} dz$ whenever $0<r<1$?

I think this has to do with Cauchy's integral formula since $2\pi i f'(0) = \int_{C_r} \frac{f(z)}{z^2}dz$. So it seems like we should have $\int_{C_r} \frac{f(z)}{z^2}dz = -\int_{C_r} \frac{f(-z)}{z^2}dz$. However, if I change $z$ to $-z$ then I get the opposite orientation so I think it should be $-\int_{-C_r} \frac{f(-z)}{z^2}dz$. How do we get this identity?

Answer 1

The substitution $z \to -z$ does not change the orientation of the curve, it corresponds to a shift $\phi \to \phi + \pi$ in the argument. Therefore $\int_{C_r} \frac{f(z)}{z^2}dz = -\int_{C_r} \frac{f(-z)}{z^2}dz$ is correct.

Alternatively you can apply Cauchy's integral formula to $g(z) = f(z) - f(-z)$, with $g'(0) = 2f'(0)$.

Answer 2

Hint: You can use Cauchy's differentiation formula: $f^{(n)}(a)=\dfrac {n!}{2\pi i}\oint_\gamma \dfrac {f(z)}{(z-a)^{n+1}}\operatorname dz$.

Let $g(z)=f(z)-f(-z)$. Then $g'(0)=2f'(0)$.

Can you finish?

Answer 3

Using holomorphic $\implies$ analytic: write $$f(z) = \sum_{n=0}^\infty a_n z^n$$ (with $a_n = f^{(n)}(0)$/n!)

Then, $$ \frac{f(z)-f(-z)}{z^2} = \frac 1{z^2}(0 + 2a_1z + \cdots) = \frac{2a_1}z + \cdots $$ and $$ \frac 1{2\pi i}\int_{|z|=r}\frac{f(z)-f(-z)}{z^2}\,dz= \hbox{residue of integrand in $0$} = \cdots $$