For $(1)$ Let $f(x)=C(\alpha)x^{\alpha}-\ln(1+x)$ We have $f'(x)=C(\alpha )\alpha x^{\alpha -1} -\frac{1}{x+1}$ Then $f'(x)\geq 0$ when $x\geq 0$ $\Leftrightarrow$ $C(\alpha )\alpha ( x^{\alpha} +x^{\alpha -1}) \geq 1$ Since $(1-\alpha )x^{\alpha} +\alpha x^{\alpha -1} \geq 1$ Choose $C(\alpha ) =\max \{ 1,\frac{1-\alpha}{\alpha} \}$ this give $\alpha \cdot C(\alpha) \geq 1-\alpha ,\alpha$ and we are done.

$\boxed{\textbf{Part (1)}}$ Since $\ln(1 + x) = x^{a}$ when $x = 0$, we just need to show that $f(x) = \frac{\ln(1 + x)}{x^{\alpha}}$ has an upper bound depending on $\alpha$ where $x > 0$ Consider $f'(x) = \dfrac{\dfrac{x^{\alpha}}{1 + x} - \ln (1 + x).x^{\alpha}}{x^{2\alpha}} = \dfrac{1 - (1 + x)\ln(1 + x)}{x^{\alpha}(1 + x)}$ Note that $f'(x) = 0$ has only one zero $x_{0} > 0$, hence $f(x)$ is increasing in $(0, x_{0})$ and decreasing in $(x_{0}, +\infty)$. Thus, $f(x) \le f(x_{0}) \quad\forall x > 0$, and we just pick $C(\alpha) \ge f(x_{0})$ and we're done. $\boxed{\textbf{Part (2)}}$ Apply the result in Part (1), we obtain $\left|\ln\left(\left|\frac{z_{1}}{z_{2}}\right|\right)\right|\le\ln\left(1 + \left|\frac{z_{1} - z_{2}}{z_{2}}\right|\right)\le C(\alpha)\left|\frac{z_{1} - z_{2}}{z_{2}}\right|$ and $0\le \ln\left(1 + \left|\frac{z_{2} - z_{1}}{z_{1}}\right|\right)\le C(\alpha)\left|\frac{z_{2} - z_{1}}{z_{1}}\right|$ since $1 + \left|\frac{z_{1} - z_{2}}{z_{2}}\right| \ge 1$ and $1 + \left|\frac{z_{2} - z_{1}}{z_{1}}\right| \ge 1$. Thus, the result follows as desired.