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)
 A: 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.
A: Since $|z|$, we can write $$ z= \cos\theta+ \sin\theta$$
Now we can derive useful properties for such $z$:
$1+z = 1 + \cos\theta + i\sin\theta = 1 + 2\cos^2\frac{\theta}{2} - 1  + 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\cos\frac{\theta}{2}(\cos\frac{\theta}{2} + i \sin\frac{\theta}{2})$
Thus, $|1+z| = 2|\cos\frac{\theta}{2}|$. Besides,
$$1 + z^2-z = 1 + \cos2\theta+ i\sin2\theta - (\cos\theta + i\sin\theta) = $$
$$2\cos\theta(\cos\theta + i\sin\theta) - (\cos\theta + i\sin\theta) = (2\cos\theta-1)(\cos\theta + i\sin\theta)  $$
Then, 
$$| 1 -z + z^2| = |2\cos\theta -1| = \left|4\cos^2\frac{\theta}{2} -3 \right| $$
If you define $x=\cos\frac{\theta}{2}$, with $x \in [-1,1]$, then we have:
$$|1+ z|+ |1 - z + z^2| = 2|x| + |4x^2 -3| = f(x)$$
Whose graph is easy to draw. Just divide the interval of $x$ in pieces such that the expressions inside the modules have constant sign.  To make it easier, note that $f(x)$ is even. Therefore, you can do:
$$x \in \left[0,\frac{\sqrt3}{2}\right] \Rightarrow f(x) = -4x^2 +2x +3$$ 
$$x \in \left[\frac{\sqrt3}{2},1 \right] \Rightarrow f(x) = 4x^2 +2x -3$$
Which gives $f(-\frac{1}{4})=\frac{13}{4}$ for the maximum and $f(\frac{\sqrt3}{2}) = \sqrt3$ for the minimum.
