Question 24
Prove that there exists a nonzero complex number $c$ and a real number $d$ such that
$$
\left|\left|\frac{1}{1+z+z^{2}}\right|-\left|\frac{1}{1+z+z^{2}}-c\right| \right|=d
$$
for all $z$ with $|z|=1$ and $1+z+z^{2} \neq 0$. (Here, $|z|$ denotes the absolute value of the complex number $z$, so that $|a+b i|=\sqrt{a^{2}+b^{2}}$ for real numbers $a, b$.)
$$
\left|\left|\frac{1}{1+z+z^{2}}\right|-\left|\frac{1}{1+z+z^{2}}-c\right| \right|=d
$$
for all $z$ with $|z|=1$ and $1+z+z^{2} \neq 0$. (Here, $|z|$ denotes the absolute value of the complex number $z$, so that $|a+b i|=\sqrt{a^{2}+b^{2}}$ for real numbers $a, b$.)