Suppose a function maps a region conformally show that the area... Suppose a function $w=f(z) \in H(\Omega)$ maps a region $\Omega$ conformally onto a region $\Omega'$: show that the area of $\Omega'$ is given by $$\iint_\Omega |f'(z)|^2 \, dx \, dy$$
I am trying to find the connection between mapping conformally and the conclusion but I cannot seem to find the correlation. 
 A: With the caveat of John Hughes, the property is simply a particular case of the multivariable change of variable theorem. Namely, if $f=u+iv$:
$$Jf(z) = \left|\matrix{u_x&u_y\cr v_x&v_y}\right|.$$
A: \begin{align}
z & = x+iy \\[8pt]
f(z) & = u+iv \\[8pt]
& x,y,u,v \in\mathbb R \\[8pt]
df & = f'(z_0)\,dz \\[8pt]
du + i\,dv  & = df = f'(z_0)\,dz = (a+ib)\,dz \\[8pt]
& = (a+ib)(dx + i\,dy) \\[8pt]
& =  (a\,dx-b\,dy) + i(a\,dy + b\,dx)
\end{align}
Therefore
$$
\begin{bmatrix} du \\ dv \end{bmatrix}
= \left[ \begin{array}{rr} a & -b \\ b & a \end{array} \right]
\begin{bmatrix} dx \\ dy \end{bmatrix}.
$$
Multiplying by a matrix has the effect of multiplying areas in the domain by the absolute value of the determinant to get the corresponding areas in the codomain.  The determinant is
$$
a^2 + b^2 = |f'(z_0)|^2.
$$
Therefore the element of area is $|f'(z_0)|^2\,dx\,dy$.
A: Hint: Try a change of variable.
You know that the area of $\Omega'$ is given by $$
\iint_{\Omega'}\,dxdy
$$
and that $\Omega'=f(\Omega)$ (I am assuming that when you say "conformally", you mean that $f$ is injective and holomorphic, as you need injectivity for the formula to hold, as the comments above show). 
