The value of derivative at zero for holomorphic function on unit disk [duplicate]

Let $\mathbb{D}$ be the unit disk of complex plain. Let $f:\mathbb{D}\rightarrow\mathbb{C}$ be a holomorphic function on $\mathbb{D}$.
Note the by Cauchy's formula, for $0<r<1$, $$2\pi if'(0)=\int_{|\zeta|=r}\frac{f(\zeta)}{\zeta^2}d\zeta$$ On the other hand, change $\zeta$ into $-\zeta$, we have $$2\pi if'(0)=\int_{|\zeta|=r}\frac{-f(-\zeta)}{\zeta^2}d\zeta$$ Summing up, $$2f'(0)=\frac{1}{2\pi i}\int_{|\zeta|=r}\frac{f(\zeta)-f(-\zeta)}{\zeta^2}d\zeta$$ Let $d=sup\{|f(z)-f(w)|:z, w\in\mathbb{D} \}$, then the above integral formula yields $$2|f'(0)|\le \frac{d}{r}$$ Taking $r\rightarrow 1$, we have $$2|f'(0)|\le d$$

The main problem is to find the condition that equality holds.
It says that the equality holds iff $f(z)=a_0+a_1z$ where $a_0, a_1$ are constants.
I wanna show the only if direction and I tried to set $$f(z)=a_0+a_1z+a_2z^2+...,\quad h(z)=\frac{f(z)-f(-z)}{z}$$ and using maximum principle ($h(0)=2f'(0)=d$, $|h(z)|\le d/r$ on $|z|=r\quad\forall 0<r<1$) to get $h$ is a constant function so that a_3=a_5=a_7=...=0

But I cannot find similar things for a_2, a_4, a_6.. Can you help me?