Deduce Poisson's integral formula from the mean value theorem This question is from the book Complex Analysis by Stein:

Let $u$ be a harmonic function in the unit disc that is continuous on it closure. Then, deduce Poisson's integral formula: $$u(z_{0}) = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-\left\lvert z_{0}\right\rvert^{2}}{\left\lvert e^{i\theta}-z_{0}\right\rvert^{2}}\,u\!\left(e^{i\theta}\right)d\theta$$ for $\left\lvert z_{0}\right\rvert<1$.
Show that if $z_{0}=re^{i\phi}$ then, $$\frac{1-\left\lvert z_{0}\right\rvert^{2}}{\left\lvert e^{i\theta}-z_{0}\right\rvert^{2}}=\frac{1-r^{2}}{1-2r\cos(\theta -\phi)+r^{2}}$$

I have read some proofs posted here and they directly proved the general result, which is really good, such as the proof here: Deriving the Poisson Integral Formula from the Cauchy Integral Formula
I understand those expert proofs, but the question in the book gives a hint which confuses me a lot. The hint:
Set $u_{0}(z)=u(T(z))$, where $T(z)=\frac{z_{0}-z}{1-\bar{z_{0}}z}$. Prove that $u_{0}(z)$ is harmonic, the apply the mean value theorem to $u_{0}$ and make a change of variables in the integral.
I am stuck in the first step, I don't really know how to show $u_{0}(z)$ is harmonic, since it is hard for me to calculate the Laplace.
Is there any way out to do this question following the hint?
Any explanations are really really appreciated!!!
Edit 1:
Okay. Thanks to the answer, I figured out how to prove $u_{0}(z)$ is harmonic, and then I proceed to next part.
Now, since $u_{0}(z)$ is harmonic in the unit disc, we apply Mean-Value property for Harmonic function to $u_{0}(z)$ at $z_{0}=0$.
$u_{0}(0)=u\circ T(0)=u(z_{0})=\frac{1}{2\pi}\int_{0}^{2\pi}u_{0}(re^{i\theta})d\theta=\frac{1}{2\pi}\int_{0}^{2\pi}u\circ T(re^{i\theta})d\theta=\frac{1}{2\pi}\int_{0}^{2\pi}u(\frac{z_{0}-re^{i\theta}}{1-\bar{z_{0}}re^{i\theta}})d\theta$, for all $0<r<1$
And I don't know how to do next. How can I get all other terms out of $u$ only learning $u(e^{i\theta})$? Also, why in the Poisson's integral formula, $r=1$? or I have to take limit $r\rightarrow 1^{-}$?
 A: Since $u$ is harmonic, it always possess a harmonic conjugate $v$, and therefore there exists an analytic function $f$ such that $\mbox{Re}[f(z)]=u(z)$. The function $T$ is a mobius transform, with a singularity at $z=1/\bar{z_0}$ and hence analytic elsewhere. So $u(T(z))=\mbox{Re}[f(T(z))]$ is harmonic since the composition of analytic functions is again analytic. 
A: $T$ is holomorphic in $B(0,|z_{0}|^{-1})$.  Therefore, if $u$ is harmonic in $B(0,1)$, it follows that $u \circ T$ is harmonic in $B(0,1)$.  
This is a general principle in complex analysis.  If $D$ is an open set, $u : D \to \mathbb{R}$ is harmonic, and $T : D \to \mathbb{C}$ is holomorphic, then $u \circ T$ is harmonic.  There are a number of ways to see this; I'll follow the elementary route.  Write $T(z) = f(z) + i g(z)$, where $f$ and $g$ are the real and imaginary parts respectively.  First, $$\frac{\partial}{\partial x}(u \circ T)(z) = u_{x}(z) f_{x}(z) + u_{y}(z) g_{x}(z), \quad \frac{\partial}{\partial y}(u \circ T)(z) = u_{x}(z) f_{y}(z) + u_{y}(z) g_{y}(z).$$  Differentiating once more, we obtain 
\begin{align*}
&\frac{\partial^{2}}{\partial x^{2}}(u \circ T)(z) = u_{xx}(z) f_{x}(z) + u_{x}(z) f_{xx}(z) + u_{yx}(z) g_{x}(z) + u_{y}(z) g_{xx}(z) \\
&\frac{\partial^{2}}{\partial y^{2}}(u \circ T)(z) = u_{xy}(z) f_{y}(z) + u_{x}(z) f_{yy}(z) + u_{yy}(z) g_{y}(z) + u_{y}(z) g_{yy}(z).\end{align*}  Recall that since $T$ is holomorphic, $f$ and $g$ are both harmonic.  Adding what we computed above and using the fact that $u_{xy} = u_{yx}$, we find
\begin{align*}
\Delta(u \circ T)(z) &= \Delta u(z) (f_{x}(z) + g_{y}(z)) + u_{x}(z) \Delta f(z) + u_{y}(z) \Delta g(z) + u_{xy}(z) (g_{x}(z) + f_{y}(z)) \\
&= u_{xy}(z)(g_{x}(z) + f_{y}(z))
\end{align*} 
Finally, since $T$ is holomorphic, $f$ and $g$ satisfy the Cauchy-Riemann equations
$$f_{x} = g_{y}, \quad f_{y} = - g_{x}.$$
Thus,
$$\Delta (u \circ T)(z) = 0.$$
Therefore, $u \circ T$ is harmonic.
