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Let $z_1,z_2$ and $z_3$ be complex numbers such that $|z_i|=1$ for $i=1,2,3$. Show that $|z_1z_2+z_2z_3+z_3z_1|=|z_1+z_2+z_3|$

This is an exercise from Jonathan S. Golan's book on Linear Algebra. To check that it is true for particular values of $z_i$, it is quite obvious that it holds when $z_i$ are the cube roots of unity.

I have checked it by computing both sides with $z_i=e^{i\theta_i}$, but the computation is extremely long and tedious. I was wondering if there is a more elegant way of proving this.

Apologies if this is a duplicate.

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4 Answers 4

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Note that $|z_1z_2z_3|=1$. Using the property that $|xy|=|x|\cdot |y|$, the LHS simplifies to $$\left|\sum_\text{cyc} z_1z_2\right|$$ $$=\left|\sum_\text{cyc} \frac{z_1z_2z_3}{z_1}\right|$$ $$=|z_1z_2z_3|\cdot\left|\sum_\text{cyc} \frac{1}{z_1}\right|$$ $$=1\cdot\left|\sum_\text{cyc} \frac{1}{z_1}\right|$$ Note that since $|z_1|=1$, we have $\frac{1}{z_1}=\overline{z_1}$, where $\overline{x}$ is the complex conjugate of $x$. Hence, the LHS is equivalent to $$\left|\sum_\text{cyc} \overline{z_1}\right|$$ $$=\left|\overline{\sum_\text{cyc} z_1}\right|$$ Using the fact that $|x|=|\overline{x}|$, this is equivalent to $$\left|\sum_\text{cyc} z_1\right|$$ Hence, the LHS is equivalent to the RHS.

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  • $\begingroup$ what does $\text{cyc}$ mean under sigma? $\endgroup$
    – p_square
    Commented Aug 14, 2021 at 7:04
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    $\begingroup$ @Algebrology It means do the summations with each permutation of the numbers. $\endgroup$
    – Alan
    Commented Aug 14, 2021 at 7:26
  • $\begingroup$ @Alan ah thanks a lot :) $\endgroup$
    – p_square
    Commented Aug 14, 2021 at 7:56
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    $\begingroup$ @Alan Abraham. Clearly $|z_1 z_2 z_3|=1$ but that does not imply $z_1 z_2 z_3=1$. So, how is it that $|\frac{z_1 z_2 z_3}{z_1} +\frac{z_1 z_2 z_3}{z_2}+\frac{z_1 z_2 z_3}{z_3} |$ equals $|\frac{1}{z_1} +\frac{1}{z_2}+\frac{1}{z_3} |$ ? $\endgroup$
    – Medo
    Commented Aug 14, 2021 at 8:12
  • $\begingroup$ @Medo You tagged the wrong Alan, not sure how you properly tag a 2 person name...but it shouldn't need tagging since its his answer. I'd answer, but I didn't understand it myself :) $\endgroup$
    – Alan
    Commented Aug 14, 2021 at 8:13
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We can rotate $z_1, z_2, z_3$ in a way so $z_1=1$. So let's suppose that $z_1=1$. The equation simplifies to: $$|z_2+z_2z_3+z_3|=|1+z_2+z_3|$$ Conjugation doesn't affect the absolute value, so we can rewrite the equation to: $$|\overline{z_2}+\overline{z_2z_3}+\overline{z_3}|=|1+z_2+z_3|$$ Conjugation of a complex unit is equal to its inverse. $$\left|\frac1{z_2}+\frac1{z_2z_3}+\frac1{z_3}\right|=|1+z_2+z_3|$$ $$\left|\frac1{z_2z_3}\right|\left|z_3+1+z_2\right|=|1+z_2+z_3|$$ And the absolute value of $\frac1{z_2z_3}$ is 1, so the equation holds.

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  • $\begingroup$ I think in the last step it should be $z_2$ not $z_1$ in the LHS. $\endgroup$
    – tmaj
    Commented Aug 14, 2021 at 8:36
  • $\begingroup$ Yes, it was a typo. Thanks! $\endgroup$
    – Hume2
    Commented Aug 15, 2021 at 14:41
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We have that $|z_1|=|z_2|=|z_3|=1$. So $|z_1 z_2 z_3|=|z_1|| z_2|| z_3|=1$ So $$|z_1 z_2+z_2 z_3+z_3 z_1|=|z_1 z_2 z_3||\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|= |\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|$$ $$=\left|\frac{\overline{z_1}}{|z_1|^2}+\frac{\overline{z_2}}{|z_2|^2}+\frac{\overline{z_3}}{|z_3|^2}\right|=|\overline{z_1}+\overline{z_2}+\overline{z_3}|$$ $$=|\overline{z_1+z_2+z_3}|=|z_1+z_2+z_3|.$$

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  • $\begingroup$ Yeah. Checked. Thanks $\endgroup$
    – Medo
    Commented Aug 15, 2021 at 1:04
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The way you proposed to derive the identity can be made a bit less tedious by not trying to meet all the multiplications "head-on". If we write $ \ z_{i} \ = \ e^{ \ i \ · \ \theta_{i}} \ \ , $ the sum $ \ z_1 + z_2 + z_3 \ $ has some (generally non-unit) modulus $ \ | \ z_1 + z_2 + z_3 \ | \ \ . $ On the Argand diagram, conjugation is equivalent to reflection about the " $ \ x-$axis", so certainly $ \ | \ z_1 + z_2 + z_3 \ | \ = \ | \ \overline{z_1} + \overline{z_2} + \overline{z_3} \ | \ = \ | \ e^{ \ -i \ · \ \theta_1} + e^{ \ -i \ · \ \theta_2} + e^{ \ -i \ · \ \theta_3} \ | $ $ = \left| \ \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \ \right| \ \ . $

But since $$ \ | \ z_1 \ · \ z_2 \ · \ z_3 \ | \ \ = \ \ | \ e^{ \ i \ · \ (\theta_1 + \theta_2 + \theta_3)} \ | \ \ = \ \ 1 \ \ , $$ we may write

$$ | \ z_1 + z_2 + z_3 \ | \ \ = \ \ | \ z_1 \ · \ z_2 \ · \ z_3 \ | \ · \ \left| \ \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \ \right| $$

$$ = \ \ \left| \ (z_1 · z_2 \ · z_3) \ · \ \left(\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right) \ \right| $$ $$ = \ \ | \ [ \ e^{ \ i \ · \ (\theta_1 + \theta_2 + \theta_3)} \ ] \ · \ (e^{ \ -i \ · \ \theta_1} \ + \ e^{ \ -i \ · \ \theta_2} \ + \ e^{ \ -i \ · \ \theta_3}) \ | $$ $$ = \ \ | \ e^{ \ i \ · \ (\theta_2 + \theta_3)} \ + \ e^{ \ i \ · \ (\theta_1 + \theta_3)} \ + \ e^{ \ i \ · \ (\theta_1 + \theta_2)} \ | \ \ = \ \ | \ z_2z_3 \ + \ z_1z_3 \ + \ z_1z_2 \ | \ \ . $$

(Except for the choice of using the "polar form", this approach is not significantly different from the other algebraic calculations from the other respondents.)

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