Really Complex!

You can show that the minimum of $\max{(|1+z|,|1+z^2|)}$ occurs on either $|1+z|=|1+z^2|$ or $|z| = 1$ (look at the local extrema of $|1+z|, |1+z^2|$ in the unit disk). In fact, the minimum occurs on $|1+z|=|1+z^2|, |z| \leq 1$ (you can maybe show this with the Maximum Modulus principle, I'm not sure). Anyway, so we want

$\sqrt{\min_{0 \leq r \leq 1, |1+z| = |1+z^2|}{1+2 r \cos(\theta) + r^2}}$ .

where we wrote $z = r e^{i \theta}$ . The constraint $|1+z|=|1+z^2|$ reduces to a quadratic in $\cos{\theta}$ (whose coefficients involve $r$ ), and we choose the solution with the minus sign (because the one with the plus sign is $\geq 2$ ). So we want

$\min_{0 \leq r \leq 1}{r^2+\frac{1}{2} \left(1-\sqrt{-4 r^4+12 r^2+1}\right)+1}.$

Taking the derivative and setting it to zero, we get $\sqrt{-4 r^4+12 r^2+1}=3-2 r^2$ , so equivalently we want to minimize $2 r^2$ where $-4 r^4+12 r^2-\left(3-2 r^2\right)^2+1=0, r=0, r=1$ . We get the minimum is $\sqrt{3-\sqrt{5}}$ .