<s>[INST] Solving a complex number inequality involving absolute values. Here is the relevant paragraph (from "Complex numbers from A to Z" by Titu Andreescu and Dorin Andrica) :

Original question : How does $\left | 1+z \right |=t$ imply $\left | 1-z+z^2 \right |=\sqrt{\left | 7-2t^2 \right |}$?
(I checked for $z=i$ , it seems it is wrong ...)
EDIT: It seems it is indeed wrong. So , how can I prove the inequality?(perhaps even the lower and upper bounds need to be changed)
 [/INST] I propose the following. As $|z|=1$ we have $z^{-1}=\bar{z}$. Then, defining $x=Re(z)$, you get :
$$|1-z+z^2|=|z| \times |z+z^{-1}-1| = |z+\bar{z}-1|=|2x-1|$$
Moreover, $|1+z|^2=(1+z)(1+\bar{z})=1+|z|^2+z+\bar{z})=2(1+x)$. Hence :
$$|1+z|+|1-z+z^2|=\sqrt{2(1+x)}+|2x-1|:=f(x)$$
For $x\in[-1,\frac{1}{2}]$, $f(x)=1+\sqrt{2(x+1)}-2x$, hence $f'(x)=\left(\frac{1}{\sqrt{2(x+1)}}-2 \right)$. This is positive from $x=-1$ to $x=-\frac{7}{8}$ and then negative till $x=\frac{1}{2}$.
Then, for $x\in[\frac{1}{2},1]$, $f(x)=\sqrt{2(x+1)}+2x-1$, so $f$ is increasing.
Thus, we have $min(f(-1),f(\frac{1}{2})) \leq f(x) \leq max\left(f(-\frac{7}{8}),f(1)\right)$. Yet $f(-1)=3$, $f(1)=3$.
Yet, $f(-1)=f(1)=3$, $f(\frac{1}{2})=\sqrt{3}$ and $f(-\frac{7}{8})=\frac{13}{4}$.
So $\boxed{\sqrt{3} \leq |1+z|+|1-z+z^2| \leq \frac{13}{4}}$, which is not at all the wanted results. ^^ I'm going to do some numeric tests.
Edit : so, after a few test with mathematica, if we take $z=\frac{1+i\sqrt{3}}{2}$ (corresponding to my minimum $x=\frac{1}{2}$), we have indeed the value $\sqrt{3}$, which is smaller than $\sqrt{7/2}$, so I guess there is a problem in the exercise.
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