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 $X$.

Let $\Gamma'$ be the family of their images under $\phi$.

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

Proof
Let $\rho'$ be a conformal metric on $Y$ in the sense of the definition of extremal length, given in local coordinates as:
 * $\rho' \left({z}\right) \left\vert{\mathrm d z}\right\vert$

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:
 * $\rho' \left({\phi \left({w}\right)}\right) \cdot \left\vert{\dfrac {\mathrm d \phi} {\mathrm d w} \left({w}\right)}\right\vert \cdot \left\vert{\mathrm d w}\right\vert$

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

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

and hence:
 * $L \left({\Gamma, \rho}\right) = L \left({\Gamma', \rho'}\right)$

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

It thus follows from the definition of extremal length that:
 * $\lambda \left({\Gamma}\right) \ge \lambda \left({\Gamma'}\right)$

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