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: 
The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.

The second method is much better, however you seem to misplaced your $-$ signs. You should get
$$|(2x - 1) \color{red}+ 2iy| = \sqrt{4x^2 - 4x \color{red}+ 1 + 4y^2} = \sqrt{5 - 4\cos \theta},$$
which has a maximum of $\sqrt{9} = 3$.

This could also be achieved through use of the triangle inequality. Consider
$$|2z - 1| \le |2z| + |-1| = 2|z| + 1 = 3,$$
with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.
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, $$\sup\_{|z| = 1} |2z - 1| = 3.$$