let $f:\mathbb {H}\to\mathbb {H}$ holomorphic, show $\Bigg|\frac{f(w)-f(z)}{f(w)-\overline{f(z)}}\Bigg|\leq\bigg|\frac{w-z}{w-\overline{z}}\bigg|$? let $f:\mathbb {H}\to\mathbb {H}$ be holomorphic. Show  that 
$\Bigg|\frac{f(w)-f(z)}{f(w)-\overline{f(z)}}\Bigg|\leq\bigg|\frac{w-z}{w-\overline{z}}\bigg|$  for all $w,z$  in $\mathbb{H}=\big \{z\in\mathbb{C}:Im z>0\big \}$
when does equality hold??
i am thinking of composing $f$ with some automorphism of unit disk .
 A: For every $a \in \mathbb{H}$ the map
$$\phi_{a} \colon z\mapsto \frac{z-a}{z- \bar{a}}$$ is an diffeo from $\mathbb{H}$ to $D$.  Therefore, for any $f\colon \mathbb{H} \to \mathbb{H}$ and any $z \in \mathbb{H}$ the map 
$$\phi_{ f(w)} \circ f \circ \phi_{w}^{-1}\colon D \to D$$ takes any $\phi_w(z)$ to $\phi_{f(w)}(f(z))$. Now, for $z = w$ this means 
$f$ takes $0 = \phi_{w}(w)$ to $\phi_{f(w)}( f(w)) = 0$. We can apply now the Schwarz lemma and conclude that 
$$|\phi_{f(w)}(f(z)) | \le |\phi_w(z)|$$ for all $z\in \mathbb{H}$.
If we have equality for some distinct $z$, $w$ then we have equality for all $w$, $z$ in $\mathbb{H}$, and, moreover, $f$ is a diffeo of $\mathbb{H}$. 
A: Hint. Consider the Cayley transformation $w=\dfrac{z-i}{z+i}$ which maps the upper-half plane $\mathbb{H}$ onto the unit disk $\mathbb{D}=\left\{|w|<1\right\}$ and use the Schwarz–Pick theorem.
A: This wikipedia page explains what you are looking for with details. Essentialy you need to use the Caley transform to reduce the problem to the unit disk and there you use Schwarz's lemma to prove it. 
For the equality case you just need to keep track of the compositions of the maps of the previous part and exploit the fact that you know when equality holds for Schwarz's lemma. 
A: To show that the function ϕ is a bi-analytic map onto the unit disk D, we need to show that it has the following properties:
ϕ is a one-to-one function from H to D.
ϕ is onto, i.e., every point w in D has a preimage in H.
ϕ is holomorphic, i.e., it is complex differentiable everywhere in H, and its derivative is nonzero.
To show that ϕ is one-to-one, suppose that ϕ(z1) = ϕ(z2) for some z1, z2 in H. Then we have:
(z1 - i)/(z1 + i) = (z2 - i)/(z2 + i)
Cross-multiplying and simplifying, we get:
z1 + i z2 + i = z2 + i z1 + i
which implies:
z1 = z2
Therefore, ϕ is one-to-one.
To show that ϕ is onto, let w be any point in the unit disk D. We want to find a point z in H such that ϕ(z) = w. We can write w in polar form as w = re^(iθ) for some real number r between 0 and 1 and some angle θ. Then we can solve for z in terms of w as follows:
w = (z - i)/(z + i)
w(z + i) = z - i
wz + wi = z - i
z(1 - w) = i(1 + w)
z = i(1 + w)/(1 - w)
Note that since |w| < 1, we have |1 - w| > 0, so z is in H. Moreover, we have:
ϕ(z) = (z - i)/(z + i) = [(i + w)/(1 - w) - i]/[(i + w)/(1 - w) + i]
= (i + w - i(1 - w))/[(i + w) + i(1 - w)]
= w
Therefore, ϕ is onto.
To show that ϕ is holomorphic, we can compute its derivative:
ϕ'(z) = d/dz[(z - i)/(z + i)]
= [(z + i)(1) - (z - i)(1)]/(z + i)^2
= -2i/(z + i)^2
It follows that ϕ is complex differentiable everywhere in H, and its derivative is nonzero (since the denominator is nonzero for all z in H). Therefore, ϕ is holomorphic.
Since ϕ is one-to-one, onto, and holomorphic, it is a bi-analytic map from H to D, as required.
A: now after that we want to show dϕ/dz never vanishes on H!
To show that dϕ/dz never vanishes on H, we can compute:
dϕ/dz = d/dz[(z-i)/(z+i)] = [(z+i)(1) - (z-i)(1)]/(z+i)^2 = -2i/(z+i)^2
Note that the denominator (z+i)^2 is never zero on H, since Im(z) > 0 and Im(i) = 1 > 0. Therefore, the derivative dϕ/dz is never zero on H.
in the last part Extend the domain of definition of ϕ to include the real line, R.
What is the image under ϕ of R? I.e. what is ϕ(R)?
we show that ϕ is a “conformal mapping” (analytic
and non-vanishing derivative)
To extend the domain of definition of ϕ to include the real line R, we can use the fact that ϕ is an even function, i.e., ϕ(-z) = ϕ(z) for all z in H. Therefore, we can define ϕ on R by setting:
ϕ(x) = ϕ(ix) = (ix - i)/(ix + i) = (1 - x)/(1 + x)i
for all real numbers x.
To find the image under ϕ of R, we can simply evaluate ϕ(x) for x in R:
ϕ(x) = (1 - x)/(1 + x)i = (-x + 1)/(x + 1)i
So the image of R under ϕ is the imaginary axis in the complex plane.
To show that ϕ is a conformal mapping, we need to show that it is both analytic and has a non-vanishing derivative.
We have already shown that ϕ is analytic on H, and it can be extended to be analytic on R as well. To show that ϕ has a non-vanishing derivative, we can compute:
dϕ/dz = -2i/(z+i)^2
Note that this derivative is never zero on H or R, since the denominator is never zero on these sets. Therefore, ϕ is a conformal mapping from H ∪ R to the complex plane.
