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: 
Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue). [![](https://i.stack.imgur.com/IWU9cm.png)](https://i.stack.imgur.com/IWU9cm.png)

By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.

Thus, $$\sup\_{|z| = 1} |2z - 1| = 3.$$
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 $z=-1$).