simple sup norm for a function on a unit circle From my understanding , the supremum norm would just be the maximum value  
So if I have a point z that lies on the unit circle $|z|=1$
Question:
what would the $\sup_{|z|=1} |2z-1|$  be ....
Is it just the maximum value of $(2z-1)$ for all points on the unit circle?
I attempted it in to ways, not sure if either is correct
First method
$ \sup_{|z|=1} |2z-1| = 1$ As you simply just take the $|z|=1 $ so $ |2(1)-1|=1$   I dont think this is correct
Second method:
You just let $z = x+iy$ where $|z|=1$ 
$|2(x+iy)-1| $ $= |(2x-1)-2iy| = $ $\sqrt{4x^2-4x-1+4y^2}$ so if z lies on the unit circle $x^2 + y^2 = 1 $ or $ x = \cosθ $,  $y = \sinθ$
So the sup would just be the maximum value for $\sqrt{3-4\cosθ} = \sqrt{7}?  $
Not sure what the maximum value for $\sup_{|z|=1} |2z-1|$ would be here.
 A: 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.
A: 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). 
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, $$\sup_{|z| = 1} |2z - 1| = 3.$$
A: 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 
$$\sup_{|z|=1}|2z-1|=2\cdot\frac 32=\color{red}3.$$
A: 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 $z=-1$).
