Show that $\log$ is a logarithm on $\mathbb{C}_-$ and check that it coincides with $\sum_{1}^\infty\frac{(-1)^{n-1}}n(z-1)^n$

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Let $\mathbb{C}_- = \{z\in \mathbb{C} \;|\; Re(z) > 0 \}$

Define $\log : \mathbb{C}_- \to \mathbb{C} $ given by $z = |z|e^{i\theta} \mapsto \ln |z| + i\theta$ where $\theta \in (-\pi, \pi)$

Show that $\log$ is a logarithm on $\mathbb{C}_-$ and check that in $D_1(1) \subset \mathbb{C}_-$ it coincides with $\sum_{1}^\infty\frac{(-1)^{n-1}}n(z-1)^n$

To show that $\log $ is a logarithms on $\mathbb{C}_-$, I will show that $\exp(\log(z)) = z$ where $z = |z|e^{i\theta}$

$\exp(\log(z)) = \exp(\ln |z| + i\theta) = |z|exp(i\theta) = z $

How can I show the second part? I have no idea how to show the second part