Angle of rotation and center of a complex map Let $\alpha (z)= \frac{z-1}{z+1} $ be a map on the upper half plane $\mathbb{H}$. What is its fixed point? What is the center and angle of this rotation?
I already computed its fixed point and I got $i$. It has only one fix point which lies in $\mathbb{H}$ so it is a rotation. Can anyone help me find the angle of rotation and center of this transfomation?
 A: You've already found the center of rotation, namely $i$. 
For the angle of rotation, compute the derivative $\alpha'(z) = \frac{2}{(z+1)^2}$, plug in $z=i$ to get 
$$\alpha'(i) = \frac{2}{(i+1)^2} = -i
$$
and take the argument of $\alpha'(i)$ which is $-\pi/2$. 
So $\alpha$ is a rotation of angle $-\pi/2$ with center of rotation $i$.
Edit: To address a comment, here's an explanation of the equality between infinitesmal rotation angle and the argument of the derivative. The explanation is based on a geometric understanding of complex multiplication.
A basic fact of calculus is that derivative of a function $w=f(z)$ at a point $z=z_0$ is the "best linear approximation" to $f$ at $z_0$, meaning that $f(z) = f(z_0) + L(h) + \text{(higher order terms in $h$)}$ where $L(\cdot)$ is a linear transformation of $\mathbb{C}$ and $h=z-z_0$. Intuitively this says that the infinitesmal effect of $f(z)$ near $z_0$ is the same as the effect of the linear transformation $L(h)$ (to be more fully correct, one first translates the complex plane by subtracting $z_0$, then one applies the linear transformation $L(\cdot)$, then one translates by adding $f(z_0)$, but I'll suppress those translations henceforth). 
Every linear transformation of $\mathbb{C}$ is multiplication by some number, and for $L(h)$ that number is $f’(z_0)$, hence $L(h) = f’(z_0) \cdot h.$ We can write the number $f’(z_0)$ in polar form as $$f'(z_0) = re^{i\theta} = r (\cos(\theta) + i \sin(\theta)), \qquad r = |f'(z_0)|, \,\, \theta = \text{arg}(f'(z_0))$$ Therefore $L(h) = h \cdot r \cdot (\cos(\theta) + i \sin(\theta))$. This means that the effect of $L$ in the $h$ plane is a stretch by a factor of $r = |f’(z_0)|$ composed with a rotation by the amount $\theta$. 
To summarize, the infinitesmal effect of $f(z)$ near $z_0$ is a stretch by a factor of $r = |f’(z_0)|$ composed with a rotation by the amount $\theta = \text{arg}(f’(z_0))$.
