If three complex numbers $z_k$ have modulus $1$, then $|z_1+z_2+z_3| = \left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|$ 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 $|z_1+z_2+z_3| = \left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|$.

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,
$$\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=\left|\frac{1}{R}\left(\frac{1}{\operatorname{cis}\alpha}+\frac{1}{\operatorname{cis}\beta}+\frac{1}{\operatorname{cis}\gamma}\right)\right| = T$$
Thus
$$|T| = \frac{1}{R} \tag{2}$$
It's easy to see that from $(1)$ and $(2)$ we get:
$$R=\frac{1}{R}$$
Thus
$$\frac{1}{1} = 1$$
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?
 A: 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
$$\left \vert \sum_k z_k\right \vert = \left \vert \sum_k \dfrac1{z_k}\right \vert$$
A: Let me give you a very easy solution to a pretty hard problem for your level.
Given : $z_i\overline z_i=1$ where, $1\leq i\leq 3$, can be equivalently written as $\;z_i=\dfrac{1}{\overline z_i}\;$ which gives $z_1+z_2+z_3=\dfrac{1}{\overline z_1}+\dfrac{1}{\overline z_2}+\dfrac{1}{\overline z_3}= \overline{\left(\dfrac{1}{z_1}\right)}+ \overline{\left(\dfrac{1}{z_2}\right)}+ \overline{\left(\dfrac{1}{z_3}\right)}$
$=\overline{\left(\dfrac1{z_1}+\dfrac1{z_2}+\dfrac1{z_3}\right)}$
In the last equality, we have used the distributivity of conjugate sign. You just need to use the fact that $|z|=| \overline{z}|$ to arrive at your result.
A: 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,$$\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=\left|\overline{z_1}+\overline{z_2}+\overline{z_3}\right|=|\overline{z_1+z_2+z_3}|=|z_1+z_2+z_3|.$$
A: 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.
A: $(z_1 + z_2 + z_3)/3$ 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.
