<s>[INST] Find the complex $z$ such $\max{(|1+z|,|1+z^2|)}$ is minimum find the complex $z$,such 
$$\max{(|1+z|,|1+z^2|)}$$ is minimum
My try: let $z=a+bi$,then
$$|1+z|=\sqrt{(a+1)^2+b^2}$$
$$|1+z^2|=|1+a^2+2abi-b^2|=\sqrt{(1+a^2-b^2)^2+4a^2b^2}$$
Then I can't,Thank you 
 [/INST] go further, there is two cases:
case I: $|1+z|\ge|1+z^2|$
$(a+1)^2+b^2\ge(1+a^2-b^2)^2+4a^2b^2 \iff a^4+2 a^2 b^2+a^2-2 a+b^4-3 b^2 \le 0 \iff (b^2-x_1)(b^2-x_2)\le 0, x_1=\dfrac{3-2a^2-\sqrt{D}}{2},x_2=\dfrac{3-2a^2+\sqrt{D}}{2},D=-16a^2+8a+9 \iff x_1\le b^2 \le x_2$
it is trivial that $D\ge 0$ for this case $\implies \dfrac{(1-\sqrt{10}}{4}  \le a \le\dfrac{ (1+\sqrt{10}}{4})$, 
$\min{|1+z|^2}=\min{(1+a)^2+b^2},(1+a)^2+b^2\ge (1+a)^2+x_1$, so the problem because to find:
the minimum of $f(a)=(1+a)^2+\dfrac{3-\sqrt{D}}{2}-a^2=2a+\dfrac{5-\sqrt{-16a^2+8a+9}}{2} \to f_{min}=3-\sqrt{5}, a=\dfrac{1-\sqrt{5}}{4},b^2=\dfrac{11-3\sqrt{5}}{4}$
caseII:$|1+z|<|1+z^2|$
I left this case for op. this case is a little more complex. and final answer is case I. 
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