Show that for $(z_n)_{n\in\mathbb N}$ in $\mathbb C,$ $$\limsup\limits_{n}n|z_n-1|<\infty\iff\limsup\limits_{n}|n\log(z_n)|<\infty$$ And further that $$\lim\limits_{n}n(z_n-1)=\lim\limits_{n}n\log(z_n)$$ if one of the limits exists.
There is a hint that I should use the logarithmic series $\sum_{n=1 } ^{\infty } (-\frac {1 } {k })^{k-1}(z-1)^k$ about the point $1 $. From which I can derive the inequality $|\log(z)-(z-1)|\le \frac {1 } {2 } |z-1|^2 $for $|z-1|<1$. But here I get stuck.
On
Hint. First of all $$ \limsup n|z_n-1|<\infty \quad\Longleftrightarrow\quad \{n|z_n-1|\}\,\,\,\text{bounded sequence} $$ and $$ \{n|z_n-1|\}\,\,\,\text{bounded} \quad\Longleftrightarrow\quad |z_n-1|<\frac{c}{n}, \,\,\,\text{for some $c>0$}. $$ Hence, $|z_n-1|<1/2$, for large enough $n$, i.e., $n\ge n_0$, which means that $\log\big(1+(z_n-1)\big)$ is expressible as a power seires $$ \log z_n=\log\big(1+(z_n-1)\big)=(z_n-1)-\frac{(z_n-1)^2}{2}+\frac{(z_n-1)^3}{3}+\cdots $$ and thus $$ n\log z_n=n(z_n-1)-\frac{n(z_n-1)^2}{2}+\frac{n(z_n-1)^3}{3}+\cdots $$ and hence $$ |n\log z_n-n(z_n-1)|\le\frac{n|z_n-1|^2}{2}+\frac{n|z_n-1|^3}{3}+\cdots \\ =\frac{(n|z_n-1|)^2}{2n}+\frac{(n|z_n-1|)^3}{3n^2}+\cdots \\ \le \frac{1}{2n\cdot 2^2}+\frac{1}{3n^2\cdot 2^3}+\cdots<\frac{1}{n}\left(\frac{1}{2^2}+\frac{1}{2^3}+\cdots\right)=\frac{1}{2n} $$
I will give an answer using a slightly modified form of the hint. Here log denotes the principal branch of logarithm. We have $log (z-1)=(z-1)-\frac {(z-1)^{2}} 2+\frac {(z-1)^{3}} 2\cdots$. Hence $|log(z-1)-(z-1)| \leq |z-1|^{2}+ |z-1|^{3}+\cdots =\frac {|z-1|^{2}} {1-|z-1|}$. Note that if any of the limits in the question exist then $z_n \to 1$. Hence, for $n$ sufficiently large we have $|z_n-1|<\frac 1 2$. This gives $|log(z-1)-(z-1)| < 2|z_n-1|^{2}$ for such $n$. From this it should be fairly easy to draw the desired conclusions. [If $\lim \sup n|z_n-1|< \infty$ there $|z_n-1|\leq C/n$ for some constant $C$so $n|z_n-1|^{2} \leq n(\frac C n)^{2} \to 0$ and this gives $|nlog(z_n)| \leq n|z_n-1|+2n|z_n-1|^{2}$, etc].