Range of $f(z)=|1+z|+|1-z+z^2|$ when $ |z|=1$ 
Find the Range of $f(z)=|1+z|+|1-z+z^2|$ when $z$ is a complex $ |z|=1$

I am able to get some weaker bounds using triangle inequality.
$f(z)< 1+|z|+1+|z|+|z^2|=5$
also $f(z)>|1+z+1-z+z^2|=|z^2+2|>||z^2|-2|=1$,but
these are too weak!.as i have checked  with WA
Is there an elgant solution using minimum calculus?. Substituiting $x+iy$ makes things very complex indeed.
Answer:
$[\sqrt{\frac{7}{2}},3\sqrt{\frac{7}{6}}]$
 A: Let $z=x+yi,$ where $x$ and $y$ are reals.
Thus, $x^2+y^2=1$ and
$$|z+1|+|1-z+z^2|=\sqrt{(x+1)^2+y^2}+\sqrt{(1-x+x^2-y^2)^2+(-y+2xy)^2}=$$
$$=\sqrt{2+2x}+\sqrt{(2x^2-x)^2+(1-x^2)(2x-1)^2}=\sqrt{2+2x}+|2x-1|$$ and we got a function of one variable $-1\leq x\leq 1.$
Can you end it now?
I got that the maximal value it's $\frac{13}{4}$ and occures for $x=\frac{-7}{8}.$
The minimal value it's $\sqrt3$ and occurs for $x=\frac{1}{2}.$
For the minimal value there is the following.
We need to prove that:
$$\sqrt{2x+2}+|2x-1|\geq\sqrt3$$ or
$$|2x-1|\geq\frac{1-2x}{\sqrt3+\sqrt{2+2x}},$$ which is true because
$$|2x-1|\left(\sqrt3+\sqrt{2+2x}\right)\geq|2x-1|\geq-(2x-1)=1-2x.$$
For the maximal value there is the following.
Let $x\geq\frac{1}{2}.$
Thus, $$\sqrt{2x+2}+|2x-1|=\sqrt{2x+2}+2x-1\leq\sqrt{2+2}+2-1=3<\frac{13}{4}.$$
Let $x\leq\frac{1}{2}$ and $\sqrt{2+2x}=y$.
Thus, $$\sqrt{2x+2}+|2x-1|=\sqrt{2x+2}-2x+1=\sqrt{2x+2}-2x-2+3=$$
$$=y-y^2+3=-\left(y-\frac{1}{2}\right)^2+\frac{13}{4}\leq\frac{13}{4}.$$
The equality occurs for $y=\frac{1}{2}$ or $x=-\frac{7}{8}.$
A: Since $|z|=1$ we substitute $z=e^{i\theta}$, to get:
$$
f(\theta)=|1+e^{i\theta}|+|1-e^{i\theta}+e^{2i\theta}|\\
= \sqrt{ (1 + \cos \theta)^2 + \sin^2\theta} + \sqrt{(1 - \cos\theta+ \cos2\theta)^2 + (-\sin\theta + \sin2\theta )^2)}\\
=\sqrt{|2+2\cos\theta |}+|(1-2\cos\theta)|
$$
We can denote $u=\cos\theta$ and investigate $f(u)=\sqrt{|2+2u |}+|(1-2u)|$ for $u$ between $-1$ and $1$ by simple calculus. We get
$\max f=\frac{13}{4}$ for $u=-\frac{7}{8}$, and $\min f=\sqrt3$ for $u=\frac{1}{2}$. This does not agree with the answer given but I could not find any error.
A: Note that $1-z+z^2=(1+r)(1+r^2)$ where $r$ is the cube root of unity.
Also, since $|z|=1$, $z\bar z=1$
Now rewrite the given expression as $√ (1+z)(1+\bar z) + √(1-z+z^2)(1-\bar z +\bar z^2)$
$\Rightarrow √ 1+z+\bar z+1 + √(z+r)(z+r^2)(\bar z+r)(\bar z+r^2)$
$\Rightarrow √(2+2\cos\theta) + √(z\bar z+zr+\bar zr+ r^2)(z\bar z+zr^2+\bar zr^2+ r)$
$\Rightarrow √(2+2\cos\theta) + √(2(r^2+r)z+2(r^2+r)\bar z +r+r^2+z^2+\bar z^2+4)$
$\Rightarrow √(2+2\cos\theta) + √(2(r^2+r+1)z+2(r^2+r+1)\bar z -1+(z+\bar z)^2-2z\bar z)$
$\Rightarrow √(2+2\cos\theta) + √((2\cos\theta)^2-3)$
$\Rightarrow (√2)\cos\frac{\theta}{2} + √(4\cos^2\theta-3)$
All that's left is minimising and maximsing this expression
which can be done with simple calculus.
Appendix:
$1-z+z^2=(z-\frac{1}{2})^2+\frac{3}{4}=0$
$\Rightarrow z-\frac{1}{2}=±i√\frac{3}{2}$
$\Rightarrow z=\frac{1}{2}+(±i√\frac{3}{2})$
