Invariance of Extremal Length under Conformal Mappings

Proposition
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
 * $$ \rho'(\phi(w))\cdot |\frac{d\phi}{dw}(w)|\cdot|dw|.$$

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
 * $$\frac{L(\Gamma,\rho)}{A(\rho)} = \frac{L(\Gamma',\rho')}{A(\rho')}.$$

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 $$Y$$.