Show that there existe a unique holomorphic function $f : D(0; 1) \to \mathbb{C}$ such that $f(0)=0$ and $f'(z)=\frac{1}{1-z}$

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Show that there existe a unique holomorphic function $f : D(0; 1) \to \mathbb{C}$ such that $f(0)=0$ and $f'(z)=\frac{1}{1-z}$.

I think I can develop $\frac{1}{1-z}$ and look for the coefficient of the general term.

How could I do this quesiton? Any ideas?

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Note that $z\in \mathbb D$ implies $\text { Re }(1-z)>0.$ Thus $\log (1-z)$ is holomorphic in $\mathbb D,$ where $\log$ is the principal value logarithm.

Suppose now $f$ satisfies the given hypotheses. For $z\in \mathbb D,$ define

$$g(z) = f(z) + \log (1-z).$$

Then $g$ is holomorphic in $\mathbb D$ and for all $z\in \mathbb D,$

$$g'(z) = f'(z) - \frac{1}{1-z} = 0.$$

Thus $g$ is constant in $\mathbb D$and since $g(0)=0,$ $g\equiv 0.$ Thus $f(z) = -\log (1-z), z\in \mathbb D.$