Maximum-Minimum value of sum of modulus of complex numbers

Question

Prove that

$\sqrt{7\over2} \leq |1+z| +|1-z+z^2| \leq 3\sqrt{7\over6} $

for all complex numbers with $|z| = 1$

My approach

Let t = $ |1+z|$

$t^2 = |1+z|^2 = (1+z)\cdot(1+\bar{z})$

= $1+\bar{z}+z+z\cdot\bar{z}$

= $1+ 2Re(z)+|z|^2$ = 2+2Re($z$) = $t^2$

$\Rightarrow Re(z) =$ ${t^2 - 2}\over 2$

Now for $|1-z+z^2|$

Let s = $|1-z+z^2|$

$\Rightarrow s^2 = {|1-z+z^2|}^2 = (1-z+z^2)\cdot(1-\bar{z}+\bar{z^2})$

$= 1-\bar{z}+\bar{z}^2-z+z\cdot\bar{z}-z\cdot\bar{z}^2+z^2-\bar{z}\cdot z^2+(z\cdot\bar{z})^2$

$= 1-(\bar{z}+z)+(\bar{z}^2+z^2)+ |z|^2-z\cdot\bar{z}\cdot(z+\bar{z})+|z|^2$

$= 1-2Re(z)+\bigl((z+\bar{z})^2-2z\cdot \bar{z}\bigr)+1-2Re(z)+1$

$= 1-2Re(z)+\bigl((2Re(z))^2-2\cdot 1\bigr)+1-2Re(z)+1$

$=3-4Re(z)+4\cdot \bigl(Re(z)\bigr)^2 -2$

$=1-4Re(z)+4\cdot \bigl(Re(z)\bigr)^2 $

$\Rightarrow s^2 = \biggl(2\cdot Re(z)-1\biggr)^2$

$\Rightarrow s = |2Re(z)-1|$

$s= |t^2-3|$

$\Rightarrow |1-z+z^2| = |t^2-3| $

But the correct solution says that $|1-z+z^2| = \sqrt{|7-2t^2|}$

Where did I go wrong ?

Also I want to solve this question using properties of complex number and then find maximum and minimum value of the sum as the function of $z$ and not using ($z = x+y\cdot i$)

Any help or worked out solution would be appreciated

Answer 1

The given inequality bounds in the question were wrong

The correct bounds will be $\sqrt{3}$ and $13\over{4}$

Let me continue my solution

Let t = $ |1+z|$

$t^2 = |1+z|^2 = (1+z)\cdot(1+\bar{z})$

= $1+\bar{z}+z+z\cdot\bar{z}$

= $1+ 2Re(z)+|z|^2$ = 2+2Re($z$) = $t^2$

$\Rightarrow Re(z) =$ ${t^2 - 2}\over 2$

Now for $|1-z+z^2|$

Let s = $|1-z+z^2|$

$\Rightarrow s^2 = {|1-z+z^2|}^2 = (1-z+z^2)\cdot(1-\bar{z}+\bar{z^2})$

$= 1-\bar{z}+\bar{z}^2-z+z\cdot\bar{z}-z\cdot\bar{z}^2+z^2-\bar{z}\cdot z^2+(z\cdot\bar{z})^2$

$= 1-(\bar{z}+z)+(\bar{z}^2+z^2)+ |z|^2-z\cdot\bar{z}\cdot(z+\bar{z})+|z|^2$

$= 1-2Re(z)+\bigl((z+\bar{z})^2-2z\cdot \bar{z}\bigr)+1-2Re(z)+1$

$= 1-2Re(z)+\bigl((2Re(z))^2-2\cdot 1\bigr)+1-2Re(z)+1$

$=3-4Re(z)+4\cdot \bigl(Re(z)\bigr)^2 -2$

$=1-4Re(z)+4\cdot \bigl(Re(z)\bigr)^2 $

$\Rightarrow s^2 = \biggl(2\cdot Re(z)-1\biggr)^2$

$\Rightarrow s = |2Re(z)-1|$

$s= |t^2-3|$

$\Rightarrow |1-z+z^2| = |t^2-3| $

Now we have $|1+z| = t$ and $|1-z+z^2| = |t^2-3|$

So the thing $|1+z| $ + $|1-z+z^2|$ becomes t + $|t^2-3|$, let's call it f(t)

Case 1 : if $t^2>3$

f(t) = $t+t^2-3$

To fetch maximum or minimum value, we differentiate with respect to t and equate the slope with 0 or use the formula of vertex of a parabola.

Using calculus

${d\over{dt}}(t+t^2-3)=0$

$\Rightarrow 1+2t =0$

$\Rightarrow t = {-1\over 2}$

$t^2 = {1\over 4}$ which is $< 3$

This means $t^2$ can never be greater than 3

Case 2 : $t^2\leq 3$

f(t) becomes $t + -t^2+3$

${d\over{dt}}(t-t^2+3)=0$

$\Rightarrow 1-2t =0$

$\Rightarrow t = {1\over 2}$

Clearly this is point of maxima as the parabola was inverted ($\because$ coefficient of $t^2$ was $< 0$)

So the maximum value of $t + -t^2+3$ = ${1\over 2} - {1\over 4} +3$ = ${13\over 4}$

For minimum value

the minimum value of the downward parabola is always $-\infty$

But we have modulus in the term $|t^2-3|$ which changes the sign when $t^2<3$

So, the minimum value of the expression $t +|t^2-3|$ is attained when :

$t +t^2-3 = t -t^2+3$

This gives $2t^2 = 6$ $\Rightarrow$ $t^2 =3$

$\Rightarrow$ $t = \pm \sqrt 3$

But t was |1+z| so it can't be negative, leaving us with $t = \sqrt 3$

So minimum value of the expression $t +|t^2-3| = \sqrt 3 + |3-3| = \sqrt 3$

Hence the inequality is

$\sqrt 3 < |1+z| + |1-z+z^2|< {13\over 4}$

The question is from the book "Complex Numbers from A to Z" by Titu Andreescu and Dorin Andrica Problem number 5 from 1.1 page number 13

Answer 2

Let $$F(u)=|1+z|+|1-z+z^2|, z=e^{iu}, u\in \Re$$ $$F(u)=2|\cos(u/2)|+\sqrt{[1-\cos u+ \cos (2u)]^2+[\sin u -\sin (2u)]^2}$$ $$\implies F(u)=2|\cos(u/2)|+\sqrt{(2 \cos u-1)^2}=2|\cos(u/2)|+|2\cos u-1|.$$ $F(u)$ is periodic with period $2\pi$ and it is symmetric w.r.r. t. $\pi$. So studying it for max/min for $u\in [0,\pi]$ would suffice.

When $\cos u \ge 1/2 \implies u\in[0,\pi/3]$, we can write $$F(u)=2 \cos(u/2)+2\cos u-1 \implies F'(u)= -\sin(u/2)-2 \sin (u) \implies F''(u)=\frac{1}{2}\cos (u/2)-2 \cos u$$ $$f'(u)=0 \implies u=0, f''(0)=-3/2, ~\text{so a max at}~ x=0 \implies F_{max1}=3$$ At $x=\pi/3$, $F(x)$ is continuous but not differentiable, we have $F(\pi/3)=\sqrt{3}$, this value will compete for the global min.

For $u \in (\pi/3,\pi]$,we have $F(u)=2\cos(u/2)+1-2\cos u \implies F'(x)= - \sin(u/2) + 2\sin u=0 \implies -\sin(u/2)[4\cos(u/2)-1]= 0 \implies u_0 = 2\cos^{-1}(1/4), F''(u_0)<0, ~\text{so next local max at}~ u=u_0\implies F_{max2}=f(u_0)=13/4$

Finally, $\sqrt{3}\le F(u)\le 13/4$, in the Fig. $F(u)$ below the red line marks $13/4$ (global max) and blue line marks $\sqrt{3}$ (global min).