Invariance of Extremal Length under Conformal Mappings

Theorem
Let $X,Y$ be Riemann surfaces (usually, subsets of the complex plane). Let $\phi:X\to Y$ be a conformal isomorphism between $X$ and $Y$.

Let $\Gamma$ be a family of rectifiable curves (or, more generally, of unions of rectifiable curves) in $X$, and let $\Gamma'$ be the family of their images under $\phi$.

Then $\Gamma$ and $\Gamma'$ have the same extremal length:
 * $\lambda(\Gamma)=\lambda(\Gamma')$

Proof
Let $\rho'$ be a conformal metric on $Y$ in the sense of the definition of extremal length, given in local coordinates as $\rho'(z)|dz|$.

Let $\rho$ be the metric on $X$ obtained as the pull-back of this metric under $\phi$. That is, $\rho$ is given in local coordinates as
 * $ \rho'(\phi(w))\cdot |\dfrac{d\phi}{dw}(w)|\cdot|dw|$

Then the area of $X$ with respect to $\rho$ and the area of $Y$ with respectto $\rho'$ are equal by definition:
 * $A(\rho')=A(\rho)$

Furthermore, if $\gamma\in\Gamma$ and $\gamma' := \phi(\gamma)$, then also
 * $L(\gamma,\rho)=L(\gamma',\rho')$

and hence $L(\Gamma,\rho)=L(\Gamma',\rho')$.

In summary, for any metric $\rho'$ on $Y$, there is a metric $\rho$ on $X$ such that
 * $\dfrac{L(\Gamma,\rho)}{A(\rho)} = \dfrac{L(\Gamma',\rho')}{A(\rho')}$

It thus follows from the definition of extremal length that
 * $\lambda(\Gamma)\geq \lambda(\Gamma')$

The opposite inequality follows by exchanging the roles of $X$ and $Y$.