Range of $f(z) = |1-z|+|1+z^2|$, where $z$ is a complex number If $z$ is a complex number such that $|z| = 1.$ then range of $f(z) = |1-z|+|1+z^2|$
Attempt: assuming $z=x+iy$ and $|z| = 1$ so $x^2+y^2$
so $f(x,y) = \sqrt{(1-x)^2+y^2}+\sqrt{(1+x^2-y^2)^2+4x^2y^2}$ 
could some help me, thanks 

HINT:
WLOG $z=\cos2y+i\sin2y$ where $y$ is real
$$1+z^2=1+\cos4y+i\sin4y=2\cos2y(\cos2y+i\sin2y)$$
$$\implies|1+z^2|=2|\cos2y|$$
$$1-z=1-\cos2y-i\sin2y=2\sin^2y-2i\sin y\cos y=-2i\sin y(\cos y-i\sin y)$$
$$\implies|1-z|=2|\sin y|$$
As $\cos2y=1-2\sin^2y=1-2s^2,$ writing $|\sin y|=s$
Case$\#1:$ For $2s^2\le1\iff0\le s\le\dfrac1{\sqrt2}\  \ \ \ (1)$
$$|1+z^2|+|1-z|=2(1-2s^2+s)=\dfrac{9-(4s-1)^2}4$$
By $(1),-1\le4s-1\le2\sqrt2-1\implies1\le(4s-1)^2\le(2\sqrt2-1)^2$
Can you take it from here?
Case$\#2:$ For $2s^2>1$
