I am reading a book on Riemann surfaces. The section I am currently reading is trying to define the hyperbolic metric on the punctured disk $D\backslash\{0\}$ in $\mathbb C$.
First let $H$ be the upper half-plane with the hyperbolic metric $\frac{1}{y^2}dzd\bar z$, then use the map $$z\mapsto e^{iz}$$ to induce a metric on $D\backslash\{0\}$. The exact words are
Consider now the map $$z\mapsto e^{iz}=w$$ of $H$ onto $D\backslash\{0\}$. This local homeomorphism (which is actually a covering) induces on $D\backslash\{0\}$ the metric $$\frac{1}{|w|^2(\log|w|^2)^2}dwd\bar w$$ with potential $\frac{1}{4}\log\log|w|^{-2}$...
However my own calculation gives a different answer:
Suppose the induced metric is $\lambda^2(w)dwd\bar w$. Since $F:z\mapsto e^{iz}$ is locally invertible, we find a sufficiently short path $\gamma$ in $D\backslash\{0\}$ such that there exist $U\subset H, V\subset D$ such that $F$ restricted to U is a homeomorphism onto $V$ and $\gamma\subset V$. Hence we have $$\int_{\gamma}\lambda(w)|dw|=\int_{F^{-1}\gamma}|dz|/{\rm Im} z$$ Let $w=Fz=e^{iz}$, then $$\int_{F^{-1}\gamma}\lambda(Fz)|F'(z)||dz|=\int_{F^{-1}\gamma}|dz|/{\rm Im} z$$ This must hold for all $\gamma$, hence $$\lambda(Fz)|F'(z)||dz|=|dz|/{\rm Im} z\implies\lambda(Fz)=\frac{1}{|F'(z)|{\rm Im}z}$$ Note that $Fz=w,\ F'(z)=ie^{iz}=iw,\ {\rm Im}z=-\log|w|$, whence $$\lambda(w)=\frac{1}{-|w|\log|w|}$$ and the metric should be $$\lambda^2(w)dwd\bar w=\frac{1}{|w|^2(\log|w|)^2}dwd\bar w$$ which is different from the one in the book (a missing exponent 2 of $|w|$.
So where did I go wrong? Or is the book wrong?