An algebra problem by Sandeep Bhardwaj

Since $|z|=1$ , we can write $z=e^{ix}$ :

$|1+z|+|1-z+z^2|=|1+z|+|z-e^{\frac13i\pi}||z-e^{-\frac13i\pi}|$

$=|1+e^{ix}|+|e^{ix}-e^{\frac13i\pi}||e^{ix}-e^{-\frac13i\pi}|$

$=|e^{\frac12ix}| |e^{\frac12ix}+e^{-\frac12ix}|$ $+ |e^{i(\frac12x+\frac16\pi)}| |e^{i(\frac12x-\frac16\pi)}-e^{-i(\frac12x-\frac16\pi)}| |e^{i(\frac12x-\frac16\pi)}| |e^{i(\frac12x+\frac16\pi)}-e^{-i(\frac12x+\frac16\pi)}|$

$= 2|\cos(\frac12x)| + 4|\sin(\frac12x-\frac16\pi)||\sin(\frac12x+\frac16\pi)|$

$= 2|\cos(\frac12x)| + 2|\cos(\frac13\pi)-\cos x|$

$= 2|\cos(\frac12x)| + 2|\frac12-2\cos^2 (\frac12x)+1|$

$= 2|\cos(\frac12x)| + |3-4\cos^2 (\frac12x)|$

Setting $y=|\cos(\frac12x)|$ , we are to maximise $2y+|3-4y^2|$ with $y\in [0,1]$ . For $y\in [\frac12\sqrt3,1]$ , the expression equals $4y^2+2y-3$ which takes its maximum of $3$ for $y=1$ . For $y\in [0,\frac12\sqrt3]$ , the expression equals $-4y^2+2y+3$ which takes its maximum of $\frac{13}4$ at $y=-\frac{2}{2\cdot -4}=\frac14$ . Hence the answer is $\frac{13}4=\boxed{3.25}$ .

First, I think the answer is 3.25. I solved this problem using definition of modulus. $|1+z| + |1 - z + z^2|$ and $|z| = 1$ $|(x + 1, y)| + |(x^2 - y^2 -x + 1, 2xy - y)|$ $\sqrt{x^2 + 2x + 1 + y^2} +|(x^2 - (1 - x^2) -x + 1, y*(2x - 1))|$ $\sqrt{1 + 2x + 1} + |2x^2 - x, y*(2x - 1)|$ $\sqrt{2x + 2} + \sqrt{(x * (2x - 1))^2 + (y * (2x -1))^2}$ $\sqrt{2x+2} + \sqrt{(2x - 1)^2 * (x^2 + y^2)}$

$\sqrt{2x+2} + |2x - 1|\sqrt{1}$ $\sqrt{2x+2} + |2x - 1|$

From $-1$ to $0.5$ the norm is $1-2x$ . Completing the square: $1 - 2x + \sqrt{2x + 2} = 0$

$(2x + 2) - \sqrt{2x + 2} = 3$

$(2x + 2) - \sqrt{2x + 2} + 0.25 = 3.25$

$(\sqrt{2x+2} - 0.5)^2 = 3.25$

This has vertex at $\sqrt{2x+2} = 0.5$

Solving for $x$ :

$2x + 2 = 0.25$

$2x = -\frac74$

$x = -\frac78$

Plugging into equation:

$\sqrt{2* -\frac78 + 2} + 1 - 2(-\frac78)$

$\sqrt{2-\frac74} + 1 + \frac74$

$\sqrt{\frac14} + \frac{11}{4}$

$\frac12 + \frac{11}{4} = \frac{13}{4} = 3.25$

From $0.5$ to $1$ it is $2x -1$ . We can analyze $2x - 1 + \sqrt{2x+2}$ . By similar analysis we find that it has a minimum at -0.5 and is increasing as x increases, so we only need to evaluate it as x = 1. That value is 3. So 3.25 is maximum over range -1 to 1.