Equivalence of conformal structure and complex structure:
1).From complex structure to conformal structure is easy. Given a complex structure, we have $ T_p M \simeq \mathbb{C}$. With complex number $i $ on tangent space, there is a natural orthogonality $<1,i>=0$.The orthogonality property is independent of the chart chosen because the transition map is holomorphic (Rienmann Cauchy equation). So the inner product structure is well-defined at the tangent spaces of every point of M which yield a Riemannian metric on $M$ hence a conformal structure.
2). On the other hand , if one is given a conformal structure on $M$, suppose one metric in this conformal equivalence is $ds^2=Edx^2+2Fdxdy+Gdy^2$. One can check this can be written in complex notation as $ds=\lambda(z)|dz+\mu(z)d \overline{z}|$ for some positvie function $\lambda(z)$ and complex valued function $\mu(z)$ with $|\mu(z)|<1$. To say the coordinate map $\phi_k: U_k \rightarrow \mathbb{C}$ is conformal, it's the same as saying this diffeomorphism preserves angles:
$$ \frac{\phi_k^{\ast} d\eta^2 (u,v)}{(\phi_k^{\ast}d\eta^2(u,u))^{\frac{1}{2}}{(\phi_k^{\ast} d\eta^2(v,v))^{\frac{1}{2}}}}=\frac{ds^2 (u,v)}{(ds^2(u,u))^{\frac{1}{2}}{ (ds^2(v,v))^{\frac{1}{2}}}}
$$
where $|d\eta|$ is the Euclidean metric on $\mathbb{C}$.
Equivalently, this is to say $|dz+\mu(z)d\overline{z}|$ is proportional to $|d\eta|=|d\phi_k(z)|=|{\phi_{k}}_{z}dz+{\phi_k}_{\overline{z}}d\overline{z}|.$
Hence a solution of the Beltrami equation ${\phi_{k}}_{\overline{z}}=\mu(z){\phi_{k}}_{z}$ yield the existence of such a conformal map.
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Notice up to this point, we are only talking about the angle-preserving property of a conformal map from the perspective of a metric other than the holomorphic(analytic) property of a conformal map! But we are heading there.
Because of this angle-preserving property, one knows the coordinate map preserves complex structure. Remember it's nothing more a rotation by $\frac{\pi}{2}$ on the plane.
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$$J{\phi_{k}}_{\ast}v={\phi_{k}}_{\ast}J v
$$
However,
$${\phi_{k}}_{\ast}(J \frac{\partial}{\partial{x}})={\phi_{k}}_{\ast}(\frac{\partial}{\partial y})={u}_y +i {v}_y$$
Where $\phi_k(x,y)=u(x,y)+iv(x,y)$
Similarly,
$$J ({\phi_{k}}_{\ast} \frac{\partial}{\partial x})= J(u_x+iv_x)=i u_x- v_x$$
So
$${u}_y +i {v}_y=i u_x- v_x$$
Thus we at last reach Cauchy Riemann which means the atlas is compex analytic.