Help needed in understanding a example of complex numbers.

Question

I was reading an example from the chapter complex numbers.
Here it is:

Prove that $${\sqrt{7\over2} }≤ |1 + z|+|1 − z + z^2| ≤ 3{\sqrt{7\over6}}$$ for all complex numbers with $|z| = 1$.

Now in the solution the author takes $|z+1|=t$. Hence giving the equation ${t^2-2\over2}=\operatorname{Re}(z)$.
Now in the next step author writes $|1 − z + z^2|=\sqrt{|7-2t^2|}$.
I don't get how this very step equation arises.

I tried it using $Z\bar{Z}=|Z|^2$ and got $|1 − z + z^2|=\sqrt{7-2t^2+z^2+\bar z^2}$.

Here is the calculation part:

$|1 − z + z^2|=\sqrt{(1-z+z^2)(1-\bar z+\bar z^2)}$
$|1 − z + z^2|=\sqrt{1-z-\bar z+z^2+\bar z^2+z\bar z+z^2\bar z^2-z\bar z^2-\bar zz^2}$
$|1 − z + z^2|=\sqrt{3-2(z+\bar z)+z^2+\bar z^2}\Rightarrow \sqrt{7-2t^2+z^2+ \bar z^2}$ {using $z\bar z=1$}

Can't get it further. Am I doing right or wrong. Please help.

Answer 1

You can use $z\bar{z}=1\implies\bar{z}=1/z$ $$(1-z+z^2)(1-1/z+1/z^2)$$ $$=(z^2-z+1)^2/z^2$$ $$=(z^4-2z^3+3z^2-2z+1)/z^2$$ $$=(z+1/z)^2-2(z+1/z)+3$$ Now use $z^2+2z+1=zt$ to get $$=(t^2-2)^2-2(t^2-2)+1$$ $$=(t^2-2)(t^2-4)+1$$

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EDIT: Thanks to Dylan and Love Invariants for it!

I'm sorry I misunderstood your question. I thought you want help in proving what I showed earlier.

Note that the answer given in book, $\sqrt{7-2t^2}$ is incorrect. You can show this by simply taking $z=-1\implies t=|1+z|=|1-1|=0\implies |1-z+z^2|=\sqrt{7-2t^2}= \sqrt{7}$. Instead, $\sqrt{(t^2-2)(t^2-4)+1}=3$ gives the correct result, $|1-(-1)+1|=3$ ;)

Answer 2

I agree with the OP that $|1-z+z^2|^2=7-2t^2+z^2+\bar{z}^2$, but note that $$z^2+\bar{z}^2=(z+\bar{z})^2-2z\bar{z}=(t^2-2)^2-2=t^4-4t^2+2,$$so $$|1-z+z^2|^2=t^4-6t^2+9\implies |1-z+z^2|=|t^2-3|.$$It looks like the book's wrong.