Our teacher gave us a hard question (according to her, it is pretty hard for our level):

Given that $|z_1| = |z_2|= |z_3|=1,z \in\mathbb{C}$, prove that $|z_1+z_2+z_3| = |\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|$.

Now, the class tried for like 40 minutes to prove that, and then the teacher came up with some really complicated proof.

I sat quietly and came up with this proof: $$|z_1+z_2+z_3| = |R(\operatorname{cis}\alpha + \operatorname{cis}\beta + \operatorname{cis}\gamma)| = I$$ Thus $$|I| = R\tag{1}$$ Also, $$\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=\left|\frac{1}{R}\left(\frac{1}{\operatorname{cis}\alpha}+\frac{1}{\operatorname{cis}\beta}+\frac{1}{\operatorname{cis}\gamma}\right)\right| = T$$


$$|T| = \frac{1}{R} \tag{2}$$

It's easy to see that from $(1)$ and $(2)$ we get:



$$\frac{1}{1} = 1$$

Which finish the proof.

My teacher said that there is a mistake in my proof, but found none - she said it could not be that easy.

Is there an error in my proof? Or is it valid?

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    $\begingroup$ Taken literally, you assume that the conjectured equality holds and prove under this assumption ("thus") that $1=1$, a fact we already knew. $\endgroup$ – Hagen von Eitzen May 5 '15 at 17:21
  • $\begingroup$ $R = 1$ right? Since $z_1$, $z_2$, $z_3$ lie on the unit circle? Am I missing something? $\endgroup$ – Alex May 6 '15 at 1:40
  • $\begingroup$ I'd say your proof is completely incomprehensible. You never say what $R$ is; you write $I$ where (I think) you mean $1$; equations (1) and (2) are both taking the absolute value of something that was already the absolute value of something else (therefore rather pointless); you seem to use that $|\def\cis{\operatorname{cis}}\cis\alpha+\cis\beta+\cis\gamma|=1$, which is false. $\endgroup$ – Marc van Leeuwen May 6 '15 at 8:07
  • $\begingroup$ ...That is a pretty hard question at your level, good on you for tackling it. $\endgroup$ – kleineg May 6 '15 at 12:49
  • $\begingroup$ @HagenvonEitzen The last equation will work only when $R=1$... $\endgroup$ – Eminem May 8 '15 at 9:23

The following is a fairly easy proof.

Since $z_1,z_2,z_3$ are on the unit circle, we have:$$\frac{1}{z_i}=\overline{z_i}\qquad i=1,2,3.$$ Hence,$$\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=\left|\overline{z_1}+\overline{z_2}+\overline{z_3}\right|=|\overline{z_1+z_2+z_3}|=|z_1+z_2+z_3|.$$

  • $\begingroup$ Aa, That same timing but better typeset! I really didn't know that over line. :) $\endgroup$ – Mann May 5 '15 at 17:22
  • $\begingroup$ Ahha - finally something that I can understand. Thank you :) $\endgroup$ – Eminem May 5 '15 at 17:23
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    $\begingroup$ @Eminem Pleasure. Now you can show this to your teacher. She may be happy to see a new solution, simpler than the one she already knows. $\endgroup$ – Amitai Yuval May 5 '15 at 18:28
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    $\begingroup$ The geometric interpretation of your first equation is that as long as $z_i$ is a unit vector, taking the reciprocal of $z_i$ is exactly the same as reflecting $z_i$ in the real line (horizontal axis). So if you imagine three unit vectors given in arbitrary position, with their sum vector constructed, just take the mirror image with respect to the real axis. That will not change the length of the sum vector. I guess this is what user238209 said concisely in his answer, the map $z\mapsto\frac{1}{z}$ is an isometry on the unit circle. $\endgroup$ – Jeppe Stig Nielsen May 6 '15 at 11:00
  • $\begingroup$ The teacher said it was hard for their level. Perhaps this means, for example, they do not know of the complex conjugate. $\endgroup$ – GEdgar May 6 '15 at 13:11

Your proof is incorrect since $\vert I \vert \neq R$.

The proof is actually fairly straightforward. In fact, we can prove that if $\vert z_k \vert = 1$ for all $k \in \{1,2,\ldots,n\}$, we have $$\left \vert \sum_k z_k\right \vert = \left \vert \sum_k \dfrac1{z_k}\right \vert$$

Let $z_k = e^{it_k} = \cos(t_k) + i \sin(t_k)$. We then have $\dfrac1{z_k} = e^{-it_k} = \cos(t_k) - i\sin(t_k)$. Hence, we have $$\sum_{k=1}^n z_k = \sum_{k=1}^n \cos(t_k) + i \sum_{k=1}^n \sin(t_k)$$ $$\sum_{k=1}^n \dfrac1{z_k} = \sum_{k=1}^n \cos(t_k) - i \sum_{k=1}^n \sin(t_k)$$ Hence, $$\left\lvert \sum_{k=1}^n z_k \right \rvert = \sqrt{\left(\sum_{k=1}^n \cos(t_k)\right)^2 + \left(\sum_{k=1}^n \sin(t_k)\right)^2}$$ and $$\left\lvert \sum_{k=1}^n \dfrac1{z_k} \right \rvert = \sqrt{\left(\sum_{k=1}^n \cos(t_k)\right)^2 + \left(\sum_{k=1}^n \sin(t_k)\right)^2}$$ which therefore gives us that $$\left \vert \sum_k z_k\right \vert = \left \vert \sum_k \dfrac1{z_k}\right \vert$$

  • $\begingroup$ But why can't I say that $|I| = R$? Also, you proof is too complicated for a high school student like me :) But im sure it is correct. $\endgroup$ – Eminem May 5 '15 at 17:16
  • $\begingroup$ @Eminem take for counterexample, $z_1,z_2,z_3$ as the three cubic roots of unity ($1, -\frac{1}{2}+\frac{\sqrt{3}}{2}i, -\frac{1}{2}-\frac{\sqrt{3}}{2}i$). Then $z_1+z_2+z_3=0\neq R=1$ $\endgroup$ – JMoravitz May 5 '15 at 17:18
  • $\begingroup$ @Eminem It might happen that $z_1=z_2=z_3$, in which case $I=3R$ $\endgroup$ – Hagen von Eitzen May 5 '15 at 17:19
  • $\begingroup$ @Eminem A even simpler example. Take $z_1=z_2=z_3 = 1$. We then have $z_1 = z_2 = z_3 = 1\cdot e^{0 \cdot i \pi}$, where $R=1$. We hence have $z_1 + z_2+ z_3 = 1\cdot e^{0 \cdot i \pi} + 1\cdot e^{0 \cdot i \pi} + 1\cdot e^{0 \cdot i \pi}$, whose magnitude is $3$ and not $R=1$. $\endgroup$ – Adhvaitha May 5 '15 at 17:20
  • $\begingroup$ @Eminem: user17762 is doing exactly the same thing as Amitai Yuval except for converting to one of the standard parametrizations of the unit circle, namely $e^{it}$. Notice that he proves that $\frac{1}{z} = z^*$ for $z$ on the unit circle using that representation. In this specific problem this parametrization is a detour since being on the unit circle already gives that fact directly, but in other problems it can help a lot. $\endgroup$ – user21820 May 6 '15 at 6:22

Let me give you an very easy solution to very hard problem :D

Given : $z_i\bar z_i=1$ Where, $1\leq i\leq 3$ Can be equivalently written as:

$z_i=\dfrac{1}{\bar z_i}$

Which gives $z_1+z_2+z_3=\frac{1}{\bar z_1}+\frac{1}{\bar z_2}+\frac{1}{\bar z_3}=\bar{(\frac{1}{z_1})}+\bar{(\frac{1}{z_2})}+\bar{(\frac{1}{z_3})}$


Now you can use the distributivity of conjugate sign and the fact that $|z|=|\bar{z}|$

To arrive at your result.


Where is the error in @Eminem 's proof?

As best as I can tell you're getting $$|T|=\frac{1}{R}$$ from one of three lines of thinking. If none of these lines of thinking apply in your case, please elaborate how you derived line (2).

A. $$(\operatorname{cis}\alpha + \operatorname{cis}\beta + \operatorname{cis}\gamma)=1$$ This is false.

or B. $$\frac{1}{(\operatorname{cis}\alpha + \operatorname{cis}\beta + \operatorname{cis}\gamma)}=\frac{1}{\operatorname{cis}\alpha}+\frac{1}{\operatorname{cis}\beta}+\frac{1}{\operatorname{cis}\gamma}$$ This is also false.

or C

@hagen-von-eitzen correctly interpreted your thinking. Assuming the theorem one is attempting to prove is a common mistake.

However, R does equal 1 because $$|z1|=|z2|=|z3|=1$$

Read through http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet.pdf from http://tutorial.math.lamar.edu/cheat_table.aspx Particularly the Common Algebraic Errors on the last page. Compare and contrast them with the related theorems on the first page and a half.

  • $\begingroup$ I initially thought that if I have something in the form of $|R(cis\alpha+ cis\beta + cis\gamma...)|$, then that equals $R$, which is apparently false. $\endgroup$ – Eminem May 6 '15 at 12:03

$(z_1 + z_2 + z_3)/3$ is the barycenter of the triangle formed by the $z_i$'s, and $z \mapsto 1/z$ acts as an isometry on the unit circle, so the barycenter of the new triangle is the same distance from the origin as the original one.

  • $\begingroup$ This is a great answer. $\endgroup$ – jwg May 6 '15 at 11:27

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