<s>[INST] 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:
$[\sqrt{\frac{7}{2}},3\sqrt{\frac{7}{6}}]$
 [/INST] 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:
$1-z+z^2=(z-\frac{1}{2})^2+\frac{3}{4}=0$
$\Rightarrow z-\frac{1}{2}=±i√\frac{3}{2}$
$\Rightarrow z=\frac{1}{2}+(±i√\frac{3}{2})$
</s>