Question: Let $h(t)$ be a smooth curve in a matrix Lie group *G*, with $h(0)=e$. Let $v \in T\_e G$. Then it holds that

$$\left.\frac{d}{dt}\left(Ad\_{h(t)}(v)\right)\right|\_{t=0} = ad\_{\left.\frac{d}{dt} h(t) \right|\_{t=0}}(v).$$

... I have been trying to figure out if this holds in a general Lie group. Any hints would be greatly appreciated.

Answer: 
Note that $\:|2z-1|=2\Bigl|z-\frac12\Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $\;A\bigl(\frac12,0\bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus 
$$\sup\_{|z|=1}|2z-1|=2\cdot\frac 32=\color{red}3.$$
The first method is definitely wrong as you suggested. This is because $|z\_1 + z\_2| \neq |z\_1| + |z\_2|$. In other words: you cannot just replace all variables inside $||$ with their "length". 

The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $\sqrt{5 - 4x})$. You do not need to represent $x$ using the angle $\theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).