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I was solving this problem 5 from book namely Complex Analysis A to Z , here we have to find range of $$ y=|1+z|+|1-z+z^2|$$ with $|z|=1$ .

enter image description here

First of all i didn't get the step $|1-z+z^2|=\sqrt{|7-2t^2|}$ .

Then i tried to solve this question by substituting $z= \cos\theta+i\sin\theta$

After simplification and plotting the graph i got $$y=|t|+\sqrt{t+3}$$ where $ t=2\cos\theta-1$ and $$\sqrt3\leq y\leq {13\over4}$$ .

So i actually got complex numbers $z=-{7\over8}\pm \frac{\sqrt15}{8}$ which gives $y={13\over4} > 3\sqrt{{7\over 6}}$ maxima we needed to prove .

So which answer is right did i make a mistake ?

There's a similar question asked so just confirm me if book is wrong and question can be merged with that ?

Also the step i mentioned earlier how author got that ?

gt6989b
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Rishi
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    Your calculation is correct (both the minimum and the maximum) – lcv May 18 '22 at 04:22
  • So book's answer is wrong ? And the step $|1-z+z^2|=\sqrt{|7-2t^2|}$in picture of book , how author got it ? – Rishi May 18 '22 at 04:26
  • Please, see my answer – lcv May 18 '22 at 04:48
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    Are you working from the 2005 edition? I have the 2014 edition of Complex Numbers... A to Z by Andreescu and Andrica. Problems 5 and 6 on pages 13 and 14 are the same, except that the inequality is shown (and solved) as $$ \sqrt3 \ \le \ |1 + z| \ + \ |1 - z + z^2| \ \le \ \frac{13}{4} \ \ . $$ The solution there is substantially similiar to what lcv shows (there are also a revised graph of the function), so it looks like you caught an error that was subsequently corrected. –  May 21 '22 at 01:23

1 Answers1

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You are correct. The mistake in the book comes from the fact that one has

\begin{align} \vert 1- z + z^2 \vert^2 &= 3-2z^* - 2z +z^2 +(z^*)^2 \\ & = (1 - z^* -z)^2, \end{align}

with $z^*$ denoting complex conjugate of $z$. Hence, using the notation of the book,

$$ \vert 1- z + z^2 \vert = \vert 3 - t^2 \vert \ . $$

Indeed $f(t) = t + \vert 3 - t^2 \vert$ satisfies

$$ \sqrt{3}\le f(t) \le \frac{13}{4} $$

when $t\in[0,2]$. I have no idea where the mistake may have originated. It also seems that this way of thought is not the simplest either.

lcv
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