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)

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.
