Question: What is the coefficient of $~\frac 1z~$ in the Laurent series expansion of $~\log\left(\frac{z}{z-1}\right)~,$ where $|z|\gt 1~?$
My work: Let $~f(z)=\log\left(\frac{z}{z-1}\right)~,~$ then
$$f(z)=\log\left(\frac{1}{1-1/z}\right)=\log 1-\log\left(1-\frac{1}{z}\right)=0-\left[-\frac{1}{z}-\frac{1}{2z^2}-\frac{1}{3z^3}-\cdots\right]=\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{3z^3}+\cdots~$$ Hence the coefficient of $~\frac 1z~$ is $~1~.$
But the answer given is $-1$.
How it is true ? Where my process is incorrect ? Any one please help.
Note: Someone may close the question by giving the reason that it is a duplicate of What is the coefficient of $\frac{1}{z}$ in the expansion of $\log(\frac{z}{z-1})$,valid in $\vert z\vert>1$?.
But the main fact is that it is a proof-verification type question. Here I want to verify where is wrong in my thinking. And also I think that my way of problem solving is different from the answer given there. So please help me.
$$f(z)=\log\left(\frac{1}{1-1/z}\right)=\log 1-\log\left(1-\frac{1}{z}\right)$$
We know the series expansion for $\log (1-w) = -w - \frac{w^2}{2}-\frac{w^3}{3} \cdots, \quad |w|<1.$
$$f(z) = -\log \left(1 - \frac{1}{z} \right) = \frac{1}{z} + \frac{1}{2z^2} + \cdots, \quad |z|>1.$$
The coefficient of $\frac{1}{z}$ is $1$.
The residue at $\infty$ is $$\text{Res}_{z=\infty}f(z)= \frac{1}{2\pi i} \oint_C f(z) \, dz$$ where the circle $C: |z|=R$ is taken $\textit{clockwise}$. Letting $w=1/z$, $$\text{Res}_{z=\infty}f(z)= \frac{1}{2\pi i} \oint_\gamma f(1/w) \left( -\frac{1}{w^2} \right)\, dw$$ where $\gamma: |w|=1/R$ is a circle taken in the counterclockwise direction.
The residue at $\infty$ is thus the residue at $0$ of $ -\frac{1}{z^2} f(1/z)$, so $$\text{Res}_{z=\infty} f(z) = -1.$$ Please see the comments for more discussion about the residue at infinity.