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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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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? $\endgroup$ Aug 14 at 7:04
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    $\begingroup$ @Algebrology It means do the summations with each permutation of the numbers. $\endgroup$
    – Alan
    Aug 14 at 7:26
  • $\begingroup$ @Alan ah thanks a lot :) $\endgroup$ Aug 14 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
    Aug 14 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
    Aug 14 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
    Aug 14 at 8:36
  • $\begingroup$ Yes, it was a typo. Thanks! $\endgroup$
    – Hume2
    Aug 15 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
    Aug 15 at 1:04

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